ENGINEERING  FOR  ARCHITECTS 


COLUMBIA  UNIVERSITY   PRESS 
SALES  AGENTS 

NEW  YORK: 

LEMCKE  &  BUECHNER 
30-32  WEST  27-ra  STREET 

LONDON: 
HUMPHREY  MILFORD 

AMEN  CORNER,  E.G. 


ENGINEERING    FOR 
ARCHITECTS 


BY 

DEWITT  CLINTON   POND,   M.A. 

INSTRUCTOR   OF    ARCHITECTURAL    ENGINEERING 
COLUMBIA    UNIVERSITY 


gotfc 

COLUMBIA  UNIVERSITY  PRESS 
1915 

All  rights  reserved 


fb 


Copyright,  1915 
BY  COLUMBIA  UNIVERSITY  PRESS 


Printed  from  type,  June,  1913 


Preface 


ARCHITECTS  often  encounter  problems  in  engineering  that 
can  be  solved  with  the  aid  of  simple  mathematics  and  a 
handbook,  published  by  a  steel  manufacturing  company.  It  is 
the  case,  however,  that  for  a  certain  problem  the  method  of  attack 
is  unknown,  and  the  architect  is  forced  to  go  to  an  engineer  or 
else  risk  failure  of  his  structure.  In  some  cases  unnecessary  cost,  is 
incurred  through  lack  of  knowledge  of  the  supporting  strength  of 
structural  members,  and  the  need  of  such  knowledge  is  felt. 

It  is  to  furnish  such  information  that  this  book  has  been 
written.  The  author  does  not  pretend  to  introduce  any  new 
methods  of  calculation,  nor  to  give  the  only  ones  that  may  be 
used.  He  is  simply  placing  at  the  disposal  of  architects  such 
information  as  will  make  possible  the  design  of  floor  beams, 
girders,  column  sections,  grillage  beams,  and  simple  roof  trusses. 
There  are,  of  course,  shorter  methods  that  experienced  engineers 
employ;  there  are  entirely  different  ways  in  which  structural 
members  may  be  designed;  but  in  case  nothing  whatever  is  known 
of  design,  it  is  the  hope  of  the  author  that  this  book  will  give 
such  information  as  will  make  the  solving  of  simple  engineering 
problems  possible.  D  c  POND 

COLUMBIA  UNIVERSITY,  1915. 


333834 


Contents 


CHAPTER  I  PAGE 

Failure   of  Beams.     Definition   of  "Moment."     Formula   for   the   maximum 

bending  moment  caused  by  a  uniform  load.     The  use  of  Handbooks  .      .          I 

\ 

CHAPTER   II 

Wood  Beams.     Moment  of  inertia.     Safe  loads.     Floor  construction  .      .      .        10 

CHAPTER  III 

Concentrated  Loads.  Reactions.  Points  of  maximum  bending  moment,  and 
shear  diagrams.  Method  of  finding  M,  the  maximum  bending  moment. 
Method  of  finding  0,  the  section  modulus  when  M  is  measured  in  foot- 
tons.  The  design  of  spandrel  beams 18 

CHAPTER   IV 

"  Built  up"  Girders.  Depth  of  girders.  Thickness  of  web  plate.  Maximum 
bending  moment.  Area  of  flanges.  Length  of  cover  plates  and  bending 
moment  diagram.  StifFeners 27 

CHAPTER  V 

Columns.  Formulas.  Assumed  section  and  safe  stress  per  square  inch. 
Dimensions.  Moments  of  inertia.  Radii  of  gyration.  Actual  safe  work- 
ing stress.  Supporting  value  of  section.  Checks 38 

CHAPTER  VI 

Grillage  Beams.  Failure  by  crushing.  Assumed  number  of  beams  in  lower 
layer.  Calculated  lengths.  Size  of  concrete  footing.  Checks.  Grillage 
for  outside  columns 45 

CHAPTER  VII 

Grillage  Foundation  on  Soil.  Size  of  concrete  footing.  Length  of  beams. 
Maximum  bending  moment  and  section  modulus  of  upper  beams.  Tests 
for  shearing  and  crushing.  Design  of  lower  layer  of  beams.  Footings  for 
corner  columns.  Cantilever  beams 53 


viii  CONTENTS 


CHAPTER  VIII 

Steel  Framing  Plan.     Record  of  calculations.     "Filling-in"  beams.     Girders. 

Trial  calculations 62 

CHAPTER  IX 

Column   Schedules.     Formulas.     Loads.     Column  sections.     Checks.     Eccen- 

ric  loads 71 

CHAPTER   X 

Graphical  Methods  of  Indicating  Forces.  Composition  of  forces.  Resultants 
and  equilibrants.  Points  of  application.  Graphical  methods  applied  to 
simple  beams 79 

CHAPTER   XI 

Simple  Truss  Problems.     Fan  truss.     Fink  truss.     French  truss.     Wind  load 

diagram,  both  ends  of  truss  fixed 87 

CHAPTER  XII 

Wind  Load    Diagram  —  One  end   free.     Tabulating  of  stresses.      Cantilever 

trusses 96 


Engineering  for  Architects 


CHAPTER  I 

Failure  of  beams.     Definition  of  "Moment."     Formula  for  the  maximum  bending 
moment  caused  by  a  uniform  load.     The  use  of  handbooks. 

THE  object  of  this  book  is  to  explain  simply,  and  plainly,  a  few 
secrets  of  engineering,  so  that  an  architect  can  decide  for  him- 
self the  size  of  beams,  girders,  and  column  sections  necessary  for 
construction.  The  solutions  of  these  secrets  are  very  simple,  and 
no  knowledge  of  mathematics  beyond  simple  algebra  is  necessary. 
If  the  architect  or  draftsman  will  follow  these  discussions  carefully, 
and  bear  in  mind  that  all  of  this  work  is  by  no  means  difficult,  he  will 
find  there  will  be  no  secrets  of  engineering  that  he  cannot  solve  for 
himself.  The  architect  is  cautioned  to  keep  from  making  a  mystery 
of  an  essentially  plain  subject. 

A  handbook,  such  as  published  by  the  Carnegie  or  Cambria 
Steel  Companies,  should  be  obtained.  The  possession  of  such  a 
book  is  absolutely  necessary.  A  slide  rule  may  be  used,  but  for  the 
simple  calculations  that  an  architect  may  find  necessary,  a  slide 
rule  is  hardly  essential. 

For  the  first  problem  we  can  take  a  beam,  properly  braced  with 
tie  rods,  which  carries  a  brick  wall  (Fig.  i).  The  span,  usually 
denoted  by  the  letter  "1,"  will  be  taken  as  twenty  feet,  the  wall 
to  be  eight  inches  thick  and  fifteen  feet  high.  The  question  is  to 
find  the  size  of  an  I-beam  which  will  carry  such  a  load.  It  will  be 
assumed  that  the  wall  is  fresh  laid  —  the  mortar  "green"  —and 
that  there  is  no  "arch  action"  caused  by  the  bonding  of  the  brick. 

The  wall  is  considered  as  distributing  its  weight  uniformly  over 
the  entire  beam,  and  the  beam  is  said  to  be  "uniformly  loaded." 
In  such  a  case,  one  half  the  load  is  carried  by  each  support.  The 
wall  is  twenty  feet  long,  fifteen  feet  high,  and  two-thirds  of  a  foot 
thick  and  the  number  of  cubic  feet  it  contains  is  given  by  the  follow- 
ing simple  multiplication,  f  X  15  X  20  =  200  cubic  feet. 


^ENGINEERING,  FOR  ARCHITECTS 


Engineers  consider  brick  as  weighing  one  hundred  and  twenty 
pounds  per  cubic  foot,  although  the  New  York  Building  Code  gives 
the  weight  as  one  hundred  and  fifteen  pounds.  The  weight  of  the 
wall  is,  therefore,  200  x  120=  24,000  pounds. 

The  weight  coming  down  on  each  pier  is  twelve  thousand  pounds, 
and  each  load  is  known  as  the  left,  or  right,  reaction.  The  left 
reaction  is  usually  denoted  as  jRi,  the  right  reaction  as 
equation  that  expresses  the  above  relations  is  RI  = 
pounds. 

In  this  case  there  are  two  things  to  be  considered:  first,  the 
stresses  set  up  in  the  beam  by  the  external  loading  —  the  brick  wall, 


and  the 

12,000 


FIGURE  i 

—  and,  second,  the  ability  of  the  beam  to  resist  these  stresses.  The 
architect  must  keep  these  two  considerations  clear  and  absolutely 
distinct  in  his  mind. 

N|  Taking  up  the  first  consideration,  engineers  generally  assume 
that  the  external  load  can  make  a  beam  fail  in  two  ways :  first,  by 
bending,  as  shown  in  Fig.  2  and,  second,  by  shearing,  as  shown  in 
Fig.  3.  Shearing  means  that  type  of  failure  caused  by  forcing  the 
particles  of  a  beam  to  slide  by  each  other.  When  a  hole  is  punched 
in  a  steel  plate  the  metal  is  "sheared"  out.  Bending  sets  up  stresses 
of  compression  and  tension  which  will  be  considered  in  the  following 
chapters. 

It  is  obvious  in  this  case  that  if  the  beam  fails  by  bending,  it  will 
fail  in  the  middle  (point  c,  Fig.  2),  ten  feet  from  either  support.  The 
tendency  to  produce  bending  in  the  beam,  at  this  point,  is  found  in 
the  following  manner. 


ENGINEERING  FOR  ARCHITECTS 


When    bending   is    considered,    the    term    "moment"    is    always 
used,  and,  at  this  point,  the  first  definitions  must  be  agreed  upon. 


15,-  few 


I3CNPIMG       •    12,  "2l=    +   J  2  O.OOO'* 

v  ter.rfpuiG  •  i  W  *  k  I  "  ^_jspj_OOO_ 
TOTAL    TSEwriNG  60,000" 

FIGURE  2 

A  moment  is  the  tendency  to  produce  rotation  about  a  point.  The 
point  is  known  as  the  "center  of  moments"  and  the  moment  itself 
is  equal  to  the  product  of  the  force,  tending  to  produce  rotation, 
multiplied  by  its  perpendicular  distance  from  the  center  of  moments, 
known  as  the  lever  arm.  In  the  language  of  college  students,  a 
moment  equals  "force  times  distance" 
or  "force  times  lever  arm." 

A  scale  is  shown  in  Fig.  4.  Either 
weight  W  or  W  can  be  taken  as  the 
force,  the  point  c  is  the  center  of  moments, 

and  the  distance  d  or  d'  is  the  perpendicular  distance  from  W  or  W 
to  c.  If  W  should  drop  to  the  position  shown  by  -the  dotted  lines, 
d  would  equal  zero  and  W  X  d  =  o.  There  would  be  no  tendency 
to  rotate  around  c  and  in  other  words  there  would  be  no  moment. 
No  force,  acting  through  a  point,  produces  a  moment. 


\ 


ENGINEERING  FOR  ARCHITECTS 


Since  moments  are  the  products  of  forces  and  distances  they 
must  be  measured  in  units  of  force  and  distance,  such  as  "foot- 
pounds" or  "inch-pounds."  If  a  force  of  five  pounds  acts  at  a 
distance  of  two  feet  from  a  point,  the  moment  will  be  ten  foot-pounds 
or  one  hundred  and  twenty  inch-pounds. 

No  bending  is  produced  unless  a  moment  exists,  hence  moments 
are  often  referred  to  as  "bending  moments"  and  are  denoted  by  the 
etter  M.  The  force  is  usually  denoted  by  the  letter  W  and  the 
perpendicular  distance,  or  lever  arm,  by  small  /.  So  M  =  Wl. 


FIGURE  4 

If  the  beam  in  Fig.  I  is  to  fail  by  bending  there  must  be  a 
moment  to  cause  this  bending.  The  moment  must  act  around  the 
center  of  the  beam  (c).  Now  the  left  reaction  (R\),  may  be  consid- 
ered as  an  upward  force  and  the  moment  caused  by  RI  equals  the 
force,  12,000  pounds,  multiplied  by  the  distance,  10  feet,  or  12,000 
pounds  x  10  feet  =  120,000  foot-pounds. 

If  RI  were  taken  from  its  position  at  the  end  of  the  beam  and 
placed,  as  shown  by  dotted  lines  in  Fig.  2a,  just  to  the  right  of  c, 
it  is  obvious  that  the  section  of  the  brick  wall  at  the  left  of  c  would 


ENGINEERING  FOR  ARCHITECTS 


tend  to  bend  the  beam  downward.  The  moment  tending  to  produce 
this  bending,  of  course,  must  be  the  product  of  a  force  multiplied 
by  a  distance.  The  force  in  this  case  is  equal  to  the  weight  of  one 
half  of  the  wall  or  \W  =  12,000  pounds.  The  lever  arm  is  the  dis- 
tance from  c  to  the  center  of  gravity  of  this  half  of  the  wall  or,  in 
this  case,  five  feet,  which  is  equal  to  J/.  The  moment  equals  12,000 
pounds  multiplied  by  5  feet  or  60,000  foot-pounds.  This  downward 
moment  exists  around  c  in  all  cases  even  with  RI  in  its  proper  position. 
With  the  reaction  acting  upward  and  one  half  the  load  acting 
downward^  the  total  moment  around  c  must  be  (Ri  X  10)  —  (J/F X  5) 
or  120,000  —  60,000  =  60,000  foot-pounds 
or  60,000  x  12  =  720,000  inch-pounds. 
It  is  obvious  that  the  bending  moment 
at  the  right  of  c  is  equal  to  that  at  the 
left  or  (R,  x  10)  -  (i/Fx  5)  -  (&  X  10) 


If  this  were  not  so,  rotation  would 
take  place  around  the  center  point  c 
which,  in  the  case  of  a  beam,  never 
happens. 

The    above    discussion    is    given     to 
show  a  method  of  finding  the  maximum    3ECT1QM  or  A 
bending   moment   caused   by   a    uniform 


FIGURE  5 


load  on  a  beam,  but,  as  a  rule,  this  method  is  too  long,  and  a 
formula  can  be  used  that  will  give  the  same  results  with  much  less 
work.  The  method  of  deriving  the  formula  is  much  the  same  as 
used  in  the  above  discussion.  jRi  =  |/F,  and  the  moment  of  RI 
around  c  equals  Ri  x  \l  or  \W  X  \l  =  \Wl.  The  downward 
moment  equals  J/Fx  \l  or  \  WL  So  the  total  bending  moment 
around  c  equals  \Wl  -  \Wl  or  f/F/  -  \Wl  =  \WL 

The  formula,  M  =  \Wl,  is  the  one  always  used  to  find  the  maxi- 
mum bending  moment  in  a  beam  carrying  a  uniform  load.  It  is, 
perhaps,  the  most  useful  formula  employed  by  engineers  and  its 
application  in  the  above  case  would  be  as  follows: 

W  =  24,000  pounds,  /  =  2o'-o"  =  240",  \Wl  =  \  X  24,000  X24O 
=  720,000  inch-pounds. 

If  the  beam  is  to  fail  by  bending  it  will  fail  because  of  this  bend- 
ing moment  of  720,000  inch-pounds  and  if  we  wish  to  select  one 
that  will  carry  the  brick  wall  a  beam  must  be  selected  that  will 
resist  this  moment.  This  is  the  consideration  spoken  of  in  the  first 


ENGINEERING  FOR  ARCHITECTS 


part  of  this  article,  namely,  the  ability  of  the  beam  to  resist  the 
stresses  set  up  by  the  external  load. 

The  bending  moment  that  a  beam  will  withstand  is  given  by 
the  formula,  M  =  S  I/c. 

M  is  the  bending  moment,  S  is  the  safe  working  stress  of  the 
material  of  which  the  beam  is  made.  In  the  case  of  steel,  S  is  always 


FIGURE  6 

taken  as  sixteen  thousand  pounds  per  square  inch  and  in  the  case 
of  wood  from  eight  to  twelve  hundred  pounds  per  square  inch  is  all 
that  is  allowed.  These  safe  working  stresses  are  given  in  the  building 
code.  /  is  the  symbol  always  used  to  denote  the  "Moment  of 
Inertia"  which  will  be  explained  later.  The  letter  c  is  the  distance 
from  the  most  remote  fiber  to  the  center  (centerof  gravity)  of  the  beam. 
Open  the  Cambria  Steel  Company's  handbook,  1912  edition,  to 
page  158,  or  an  old  Carnegie  handbook  to  page  97,  or  the  "Pocket 
Companion,"  published  by  the  Carnegie  Steel  Company,  1913  edi- 
tion, to  page  142.  In  the  second  column  from  the  left,  under  the 
heading  "Depth  of  Beam,"  a  fifteen-inch  /-beam  is  found  that 
weighs  42  pounds  (Cambria)  or  36  pounds  (Carnegie,  1913).  The 
Cambria  beam  will  be  considered  first. 


ENGINEERING  FOR  ARCHITECTS 


The  area  of  the  cross  section  is  12.48  square  inches.  The  thick- 
ness of  the  web,  "t,"  is  41/100  of  an  inch.  The  flange  is  5!  inches 
wide  and  the  moment  of  inertia  or  /  is  441.8.  Obviously,  as  the 
depth  of  the  beam  is  fifteen  inches,  the  distance  c  (Fig.  5)  is 


X  15  =  7-5-     So/A  = 


-  58.9. 


-•-             ,            ts-I-SS*         i- 

•I 

Co 

Cfo 

>  •; 

\ 

do 

IT 

,    'i 

:  *°               \- 

I/c  is  known  as  the  "section  modulus"  of  the  beam,  and  this  is 
given  in  column  8,  so  for  any  steel  beam  it  is  unnecessary  for  the 
architect  or  engineer  to  go  through 
this  simple  form  of  division. 

In  the  formula  M  =  S  I/c,  for 
any  standard  section  of  I-beam, 
channel,  or  angle,  we  have  every 
factor  known  but  the  bending 
moment,  M.  For  this  particular 
fifteen-inch,  forty-two  pound  I- 
beam  the  moment  can  be  deter- 
mined by  simply  substituting  the 
formula  as  follows:  M  =  16,000  x 
58.9  —  940,000  inch-pounds,  ap- 
proximately, and  this  is  the  bend- 
ing moment  that  the  beam  will  withstand.  As  a  rule,  M  is  known 
and  I/c,  the  section  modulus,  is  the  unknown  factor.  Such  is  the 
case  with  the  simple  beam  carrying  the  brick  wall.  M,  in  this 
case,  is  720,000  inch-pounds.  To  determine  the  size  of  the  I-beam 
required,  the  formula  would  be  written  as  follows:  M  =  720,000 
inch-pounds  =  16,000  x  I/c,  or  I/c  =  720,000/16,000.  I/c  =  45. 

Looking  down  column  8  in  the  handbook  58.9  is  the  next  value 
for  I/c  above  45,  so  a  fifteen  inch,  forty-two  pound  I-beam  is  required 
to  carry  the  wall. 

If  we  consider  the  Carnegie  handbook,  the  fifteen-inch  I-beam, 
weighing  thirty-six  pounds  per  foot,  will  have  a  section  area  of  10.63 
square  inches,  a  flange  width  of  5^  inches,  with  a  web  thickness  of 
289/1,000  of  an  inch.  The  /,  or  moment  of  inertia,  of  this  beam  is 


FIGURE  6a 


405.1,  and  as  c  =  J  X  d  =  7.5,  I/c 


7-5 


which  is  found  in 


the  column  marked  "  S  "  in  the  Carnegie  book.  "  S,"  in  the  handbook, 
denotes  the  section  modulus  or  I/c.  So  a  fifteen  inch,  thirty-six 
pound  I-beam  would  withstand  a  maximum  bending  moment  of 
M  =  16,000  X  54  =  864,000  inch-pounds. 


8         ENGINEERING  FOR  ARCHITECTS 

Following  the  calculations  given  for  the  Cambria  beam  it  is 
easily  seen  that  this  beam  will  carry  the  brick  wall. 

The  second  method  of  failure  is  due  to  shearing  and  in  this  case 
the  beam  might  fail  in  this  manner  rather  than  by  bending,  as  a 
brick  wall  is  practically  rigid  after  the  mortar  is  set. 

The  shear  is  always  greatest  at  the  supports  and  is  equal  to  the 
reactions.  Looking  at  Fig.  3  it  might  be  imagined  that  the  wall 
acts  like  a  punch  —  tending  to  punch  the  beam  between  the  two 
supports,  and  the  force  exerted  at  each  pier  is  equal  to  each  reaction. 

Therefore  the  maximum  shear  in  the  beam  equals  12,000  pounds. 

Steel  will  safely  stand  a  tensile  stress  or  pull  of  16,000  pounds 
per  square  inch,  but  its  shearing  value  is  only  9,000  pounds  per 
square  inch,  as  allowed  by  the  New  York  Building  Code,  Section  139. 

The  cross-section  area  of  the  Cambria  beam,  as  given  in  the  hand- 
book, is  12.48  square  inches.  If  the  shearing  strength  of  the  steel 
is  9,000  pounds  per  square  inch  and  there  are  12.48  square  inches, 
naturally  the  shearing  value  of  the  cross  section  is  12.48  x  9,000  = 
109,320  pounds.  The  beam  is  undoubtedly  safe. 

It  is  a  significant  fact  that  although  moments  are  measured  in 
foot-pounds  or  inch-pounds,  foot-tons  or  inch-tons,  as  the  case  may 
be,  shearing  values  are  measured  in  pounds  or  tons  only.  If  the 
architect  finds  that  he  is  trying  to  measure  moments  in  terms  of 
weights  only,  he  had  better  stop  and  consider  that  it  would  be  as 
impossible  to  do  this  as  to  measure  a  mile  with  a  quart  measure. 
The  unit  does  not  fit  the  condition. 

The  maximum  shear  is  always  found  at  the  reactions  and  when 
the  reactions  are  unequal  the  greatest  shear  is  found  at  the  greatest 
reaction. 

As  a  rule  beams  are  seldom  figured  to  withstand  shear  as  long 
beams  are  more  apt  to  fail  by  bending.  However,  in  the  case  of 
heavy  loads,  carried  over  short  spans,  failure  by  shearing  must 
always  be  considered.  The  bending  moment  caused  by  a  heavy 
load  over  a  short  span  will  be  small  as  the  maximum  moment  equals 
\Wl  and  /  in  this  case  is  small.  If  the  beam  is  designed  only  for 
the  purpose  of  resisting  this  M,  it  might  easily  be  too  light  to  resist 
the  shear. 

In  the  foregoing  problem,  a  single  case  is  given  where  a  uniformly 
distributed  load  is  considered.  Another,  and  probably  the  most 
frequent  case,  is  found  where  the  beam  is  one  of  a  series  of  "filling-in 
beams,"  as  shown  in  the  section  of  a  steel  framing  plan  (Fig.  6).  A 


ENGINEERING  FOR  ARCHITECTS 


girder  is  "framed"  between  columns  120  and  no,  and  another  be- 
tween columns  121  and  in.  Filling-in  beams,  spaced  six  feet  on 
centers,  are  framed  between  the  girders.  The  floor  area  carried  by 
each  beam  is  shown  enclosed  in  the  space  abed,  and  is  equal  to  6/-o'/ 
X  i8'-o"  =  1 08  square  feet.  Assuming  a  floor  load  of  200  pounds 
per  square  foot,  then  the  total  load  on  each  beam  is  108  x  200  =21,- 
600  pounds  =  W. 

The  span  of  the  beam  is  i8'-o"  or  216"  and  so  the  bending  moment 

.    21,600x216  . 

equals  —    — - —     -  =  583,200  inch-pounds, 

o 

M  =  583,200  =  S  I/c  =  16,000  x  lie.     lie  =  ^^  =  36.4. 

In  the  handbook  a  twelve  inch,  thirty-five  pound  I-beam  will 
be  found  having  a  section  modulus  of  38.0  which  will  be  satisfactory 
for  this  condition. 

The  architect  can  make  problems  for  himself,  such  as  assuming  a 
floor  load  of  175  pounds  or  a  span  of  20  feet  in  place  of  18  feet.  If 
the  beams  and  girders  were  framed  in  opposite  directions,  as  shown 
in  Fig.  6a,  the  beam  would  have  a  span  of  24  feet  and  therefore  a 
15"  I-beam,  weighing  55  pounds  per  foot,  would  be  necessary. 

If,  in  Fig.  6,  we  assume  that  the  girders  carry  a  section  of  the  floor 
area  enclosed  by  the  lines  ef,  fg,  gh,  and  he,  the  floor  area  will  be  1 8 
feet  by  24  feet,  or  432  square  feet.  432  X  200  =  86,400  pounds  =  W. 
\Wl  =  \  X  86,400  x  24  X  12  =  3,110,400  inch-pounds. 

M  =  3,110,400  =  16,000  x  I/c,  or 

3,110,400 
I/c  =  ^-^-     ~=  194.4. 

10,000 

Therefore,  the  girders  would  have  to  be  24"  I-beams,  weighing 
100  pounds  per  foot. 

•  The  method  of  figuring  floor  loads  as  well  as  a  little  more  theo- 
retical consideration  of  the  principles  involved  in  the  above  discus- 
sion will  be  taken  up  in  the  next  chapters.  For  the  present  it  may 
be  well  for  the  architect  to  "check  up"  the  sizes  of  beams  shown  on 
any  steel  framing  plans  that  he  may  have. 


CHAPTER   II 

Wood  beams.     Moment  of  Inertia.     Safe  loads.     Floor  Construction. 

THE  two  formulas  given  in  the  first  chapter,  M  =  S  I/c  and 
M  =  \Wl,  are  the  two  most  common  formulas  used  in  en- 
gineering work.  The  "I,"  used  in  the  first,  stands  for  the  moment 
of  inertia  of  the  cross  section  of  the  beam.  To  give  a  definition  of 
this  term  would  be  useless  as  it  would  only  cause  confusion.  For 
any  steel  beam  the  moment  of  inertia  is  given  in  the  handbooks  pub- 
lished by  the  steel  companies.  If,  however,  the  architect  should  want 
to  know  the  size  of  a  wooden  beam,  strong  enough  to  carry  a  given 
load,  he  must  find  this  factor  for  himself.  This  is  by  no  means  a 
difficult  task. 

The  cross  section  of  all  wooden  beams  is  rectangular,  and,  for 
this  kind  of  a  section,  the  formula  that  gives  the  moment  of  inertia 
is  /  =  i /i  2  bd3,  in  which  b  denotes  the  breadth  of  the  beam,  and  d 
denotes  the  depth.  So,  for  a  2"  X  10"  wooden  joist,  the  formula 
would  be  written,  /  =  1/12  X  2  X  10  X  10  X  10,  and  /,  for  this 
beam,  would  be  166.6.  This  is  all  that  is  necessary  for  an  architect 
to  know  about  moments  of  inertia. 

I/c,  the  section  modulus,  is  found  by  using  the  formula  I/c  = 
1/6  bd2.  If  the  architect  bears  in  mind  that  c  equals  one  half  of  d, 
and  that  /  =  1/12  bd9,  he  can  easily  derive  this  formula  for  himself. 
The  section  modulus  of  the  joist  can  be  determined  in  the  following 
manner:  I/c  =  1/6  X  2  X  iox  10  =  33.3.  This  checks  with  the 
fact  that  166.6  divided  by  5  equals  33.3. 

To  find  the  safe  uniform  load  that  the  joist  will  carry  over  a 
span  of  ten  feet  the  formula  M  =  \Wl  is  used: 

|  X  W  X  iox  12  =  S  I/c  =  1,200  X  33.3. 
\XW=   1,200  X  33.3/120. 

W  =  333-3  X  8  =  2,666  pounds. 

In  the  steel  handbooks,  under  the  heading  of  "Safe  Loads  in 
Pounds  for  Wooden  Beams"  or  simply  "Wooden  Beams/'  loads 
are  given  for  joists,  one  inch  thick,  for  various  spans  and  depths. 
On  page  352  in  the  Cambria  Steel  Company's  handbook,  the  safe 


ENGINEERING  FOR  ARCHITECTS        u 

load  for  i"  x  10"  wood  beam  spanning  ten  feet,  is  given  as  1,333 
pounds.  A  2"  joist  would  be  twice  as  strong  and  would  carry  a 
load  of  2,666  pounds,  which  checks  with  the  answer  given  above. 
It  is  suggested  that  the  architect  figure  the  safe  load  for  a  3"  X  12" 
beam,  having  a  safe  working  stress  of  1,000  pounds  per  square  inch, 
and  a  span  of  is'-o".  The  load  should  work  out  to  be  3  X  1,067 
or  3,201  as  given  on  page  350  of  the  Cambria  book. 

The  calculation  for  the  above  problem  is  very  simple  and  should 
take  not  more  than  three  or  four  minutes. 

Of  course  the  safe  loads  for  wooden  beams  are  given  and  it  is 
really  unnecessary  for  the  architect  to  determine  them,  but  the  calcu- 
lations serve  to  make  the  terms,  "moment  of  inertia"  and  "section 
modulus/'  familiar  ones.  The  importance  of  this  familiarity  cannot 
be  too  strongly  emphasized  as  the  process  of  rinding  the  sizes  of 
beams  and  girders  should  be  almost  automatic. 

The  real  problem  that  any  designer  has  to  solve  is  to  determine 
the  kind  and  the  magnitude  of  the  loads.  Once  these  questions  are 
settled,  a  method  of  finding  the  size  of  beam  to  carry  the  loads  should 
take  but  little  thought. 

There  are  two  kinds  of  loads  —  uniform  and  concentrated.  So 
far  we  have  only  dealt  with  the  first,  but  later  we  will  take  up  the 
problem  of  finding  beams  suitable  for  carrying  concentrated  loads. 

The  processes  of  finding  magnitudes  of  loads  will  be  taken  up 
at  once. 

There  are  two  types  of  loads,  dead  loads  and  live  loads.  The 
dead  loads  include  the  weight  of  walls,  and  weight  of  the  floor  con- 
struction. These  weights  are  usually  known  as  floor  loads  and 
wall  loads.  In  Fig.  7  the  wall  load  is  carried  on  beams  known  as 
spandrel  beams,  and  the  floor  load  is  carried  on  filling-in  beams. 
In  some  cases,  however,  the  spandrel  beams  have  to  carry  some  of 
the  floor  load  as  well  as  the  wall  load. 

The  floor  construction,  in  this  case,  is  made  of  a  six-inch  terra 
cotta  arch,  between  the  beams,  on  which  cinder  fill  is  carried.  The 
cinders  fill  in  over  the  haunches  of  the  arch  and  are  levelled  off"  at 
the  tops  of  the  /-beams.  On  the  cinders  wooden  sleepers  are  placed, 
between  which,  cement  mortar  is  laid.  The  wooden  flooring  is  nailed 
to  the  sleepers  and  a  hung  ceiling  is  suspended  from  the  lower  flanges 
of  the  /-beams. 

Under  the  heading  "Arches"  in  the  handbooks  the  weight  of 
terra  cotta  arches  is  found.  It  will  be  found  that  a  six-inch  arch  is 


12 


ENGINEERING  FOR  ARCHITECTS 


given  as  weighing  twenty-seven  pounds  per  square  foot  of  floor  area. 
Also,  under  the  heading  of  "Weights,"  the  weight  per  cubic  foot  is 
given  of  cinders,  cement,  and  wood. 

The  beams  will  be  taken  as  being  twelve  inches  deep,  spaced 
six  feet  on  centers  (Fig.  9).     The  rise  of  the   arch  will  be  an  inch 


-7z  IS  i  an.  or  ARCH 

X     >6'ATECH 


FIGURE  7 

and  a  quarter  per  foot  of  span  as  required  by  the  New  York  Build- 
ing Code.  For  a  span  of  six  feet  the  rise  (Fig.  7)  will  be  six 
times  one  and  one-quarter  or  seven  and  one-half  inches.  The  top 
of  the  arch  will  be  6"  +  7.5"  =  13.5"  above  the  bottom  flange  of  the 
/-beam  or  1.5"  below  the  top  flange,  in  case  a  15"  I-beam  is  used. 
There  will  be  a  triangular  fill  over  the  haunches  of  the  arch 
and  a  "straight"  fill  of  an  inch  and  a  half  over  the  tops  of  the  arches 
to  the  tops  of  the  beams.  All  of  this  fill  will  be  made  of  cinders 
weighing  45  pounds  per  cubic  foot.  As  is  shown  in  Fig.  8,  a  layer 

of  cinders,  one  inch  deep  and  a  foot  square, 
will  weigh  one-twelfth  of  45  pounds,  or, 
approximately,  4  pounds. 

The  filling  over  the  haunches  is  roughly 
in  the  shape  of  a  triangle,  having  a  base  of 
six  feet  and  an  altitude  of  seven  and  a  half 
inches.      This   triangle  is  equivalent   to  a 
fill   of  three    and    three-quarters 
Add    to    this   the   inch   and 
the    sum    will    be    five    and 


FIGURE  8 

A  Layer  i"  thick  and  a  foot 
square  weighs  one  twelfth  as    straight 

inches   deep. 

one-half  at  the  top  of  the    arch  and 
one-quarter  inches. 

On  top  of  the  cinders  the  sleepers  are  placed  and  the  space  be- 
tween the  sleepers  is  filled  in  with  cinder  concrete.  Engineers  con- 
sider this  portion  of  the  floor  construction  as  weighing  the  same  as 


ENGINEERING  FOR  ARCHITECTS  13 

the  cinder  fill.  To  the  5.25"  add  3.75*  and  the  sum  is  9.00".  This 
is  the  average  depth  of  the  fill  between  the  arch  and  the  rough 
flooring,  and  as  each  inch  weighs  4  pounds  the  total  weight  is  36 
pounds. 

The  average  weight  of  a  cubic  foot  of  wood  can  be  taken  as  thirty 
pounds.  A  layer,  one  inch  thick  and  a  foot  square,  will  weigh  one- 
twelfth  of  thirty  or  two  and  one-half  pounds.  There  is  an  approxi- 
mate thickness  of  wood  flooring  of  two  inches,  which  will  weigh 
five  pounds  per  square  foot  of  floor  area. 

There  still  remains  the  weights  of  the  ceiling  and  the  steel  beams 
and  girders  to  be  considered.  A  hung  ceiling  is  always  supposed 
to  weigh  ten  pounds  per  square  foot  so  this  item  is  easily  disposed 
of,  but  to  understand  the  method  of  figuring  the  weight  of  the  steel 
as  a  part  of  the  floor  load,  it  is  necessary  to  study  the  framing  plan 
(Fig.  9).  The  portion  of  the  floot  enclosed  in  the  area  a  b  c  d  is  the 
size  of  a  floor  panel,  i8'-o"  x  24'-o",  and  in  it  are  found  portions 
of  beams  and  girders,  such  as  those  that  make  up  the  average  panel. 
There  is  a  part  of  a  twenty-four  inch  /-beam  weighing  one  hundred 
pounds  per  foot.  The  length  of  the  beam  found  in  area  a  b  c  d  is 
24  feet,  so  the  total  weight  of  the  girder  will  be  24  X  100  =  2,400 
pounds.  There  are  four  filling-in  beams,  each  of  which  is  i8'-o" 
long  and  weighs  forty  pounds  per  foot.  The  total  weight  of  the  beam 
is  then  4  X  18  X  40  =  2,880  pounds.  Add  this  to  2,400  pounds  and 
the  steel  in  the  panel  will  weigh  5,280  pounds.  There  are  432  square 
feet  in  the  panel,  so  the  weight  per  square  foot  of  floor  area  of  the 
steel  will  be  5,280  -f-  432  =  12.2  pounds.  To  be  safe,  and  to  give  an 
approximate  figure  for  the  steel,  this  weight  will  be  taken  as  1 3  pounds. 

To  find  the  weight  of  a  square  foot  of  floor  construction  add 
all  the  weights  given  above. 

10  pounds  =  weight  of  ceiling. 
27  "       "  terra  cotta  arch. 

36        "       =         "       "cinder  fill. 
5        "       =         "       "  wood  floor. 
13        "       =         "       "steel 
91  =  Total  weight  of  floor. 

The  ninety-one  pounds  is  the  total  dead  load  per  square  foot 
but  there  is  still  a  live  load  to  be  considered.  The  live  load  is  always 
given  in  the  building  codes  of  the  different  cities  and  all  that  the 


ENGINEERING  FOR  ARCHITECTS 


architect  has  to  do  is  to  find  what  live  load  per  square  foot  is  re- 
quired for  the  particular  building  he  is  designing. 

The  New  York  Building  Code,  in  section  130,  gives  the  live  load 
on  floors  of  dwellings  and  apartment  houses  as  not  less  than  sixty 
pounds.  If  a  building  is  to  be  used  as  an  office  building,  the  live 
load  is  figured  as  seventy-five  pounds  on  all  floors  above  the  first, 
and  on  that  the  engineer  must  figure  for  a  live  load  of  one  hundred 


FIGURE  9 
To  determine  the  weight  of  steel  for  the  Dead  Load  these  sizes  are  assumed. 

and  fifty  pounds.  Floors  of  schools  must  be  strong  enough  to  carry 
seventy-five  pounds  per  square  foot.  A  live  load  of  ninety  pounds 
is  allowed  on  floors  of  places  of  public  assembly,  and  store  floors 
must  carry  one  hundred  and  twenty  pounds  per  square  foot. 

In  case  the  floor  construction,  given  above,  is  to  be  used  in  a 
department  store,  the  live  load  upon  it  will  be  120  pounds,  so  the  total 
weight  per  square  foot  of  floor  area  will  be  91  +  120  =  211  pounds. 
As  an  easy  figure  to  deal  with,  this  total  load  can  be  taken  as  210 
pounds. 

In  the  floor  plan  (Fig.  9)  the  weight  carried  by  each  twelve- 
inch  beam  will  be  6  X  18  X  210=  22,680  pounds.  The  bending 
moment  will  be  \Wl  =  £  X  22,680  X  18  X  12  =  612,360  inch-pounds. 
Dividing  this  by  the  safe  working  stress  of  steel — 16,000  pounds 
per  square  inch  —  we  will  get  a  value  for  I/c  of  612,360  -;-  16,000  = 
38.2.  The  section  modulus,  or  I/c,  as  given  in  the  steel  companies' 
handbooks,  for  a  i2//-i-4olb  will  be  a  little  larger  than  this. 


ENGINEERING  FOR  ARCHITECTS  15 

The  type  of  floor  arch  that  we  have  been  investigating  is  only 
one  of  many  that  are  used  in  floor  construction.  Where  it  is  desir- 
able to  plaster  directly  on  the  soffit  —  the  under  side  —  of  the  arch, 
flat  terra  cotta  arches  are  used.  The  New  York  Building  Code,  in 
section  106,  contains  the  statement  that  the  "depth  shall  not  be 
less  than  one  and  three-quarter  inches  for  each  foot  of  span,  not 
including  any  portion  of  the  depth  of  the  tile  projecting  below  the 
under  side  of  the  beam,  if  the  soffit  of  the  beam  is  straight."  This 
means  that  for  a  span  of  six  feet  the  arch  must  be  ten  inches  and  a 
half  deep  above  the  bottom  flange  of  the  beams.  If  an  absolutely 
flat  ceiling  is  desired,  to  this  ten  and  one-half  inches  the  depth  of 
the  fireproofing  on  the  lower  flange  of  the  beams  must  be  added. 
This  fireproofing  can  be  considered  as  being  an  inch  and  one-half 
thick  so  the  total  depth  of  the  arch  will  be  10.5  +  1.5  =  12". 


uuuuranooDQS 


FIGURE  10 
Flat  Terra  Cotta  Floor  Arch. 

In  the  terra  cotta  manufacturing  companies'  handbooks  the  depth 
of  arch  required  for  the  above  condition  is  given  as  ten  inches. 
Although  the  twelve-inch  depth  is  excessive  it  must  be  used  in  New 
York.  The  weight  of  a  twelve-inch  arch  can  be  taken  as  38  pounds 
per  square  foot  of  floor  area.  If  the  arch  is  sprung  between  fifteen- 
inch  beams  (Fig.  10),  the  distance  between  the  tile  and  the  wooden 
flooring  is  fourteen  inches,  approximately.  As  each  inch  of  fill 
weighs  four  pounds,  the  weight  will  be  4  x  14  =  56  pounds.  The 
flooring  and  steel  will  weigh  the  same  as  in  the  first  case,  so  the 
total  weight  will  be: 

38  pounds  =  weight  of  arch. 

56        "  *         "  "fill. 

5  "  =        "  "floor. 

^3  "  "  steel. 

112  "  "  a  square  foot  of  floor. 

120  "  =         "  "live  load. 

232  "  =  Total. 

Taking  tjjj^ej^oad  as  230  pounds  per  square  foot  of  floor  area,  the 
section  modulus  of  41.8  is  obtained.  A  fifteen  inch,  forty-two 


i6 


ENGINEERING  FOR  ARCHITECTS 


pound  I-beam  will  be  required  to  take  this  load.  As  there  is  only  a 
difference  of  two  pounds  between  this  and  the  twelve-inch  beam,  it 
will  be  practically  as  cheap  to  use  the  deeper  section. 

It  will  be  noticed  that  in  both  Fig.  7  and  Fig.  10  the  arches  are 

sprung  between  fifteen-inch  beams,  although   twelve-inch   filling-in 

beams  are  often  shown  in  the  plans.     The  reason  for  the  use  of  the 

rirt. 

ELEVATION    or 
15  CAMS 


T~   T-  T" ~^~  -TO 

in  i  Hi! 


20      18      Iff       12'      10 


SPRING   OF  ARCHES 


FIGURE  n 

deeper  beams  is  as  follows:  on  all  framing  plans  the  relative  heights 
of  steel  beams  are  shown  in  the  same  manner  as  given  in  Fig.  n. 
All  beams,  having  a  greater  depth  than  fifteen  inches,  are  shown 
"flush  top"  with  the  fifteen  inch  I-beams.  This  means  that  the 
top  flanges  of  these  beams  are  all  on  the  same  level,  usually  from  five 
to  six  inches  below  the  finished  floor  line.  All  beams  having  a 

depth  of  less  than  fifteen  inches  are 
shown  "flush  bottom"  with  the  fifteen- 
inch  beams.  As  nearly  all  arches  spring 
from  15",  12",  or  8"  filling-in  beams,  this 
arrangement  makes  it  possible  for  the 
bottom  of  the  arches  to  be  on  the  same 
level.  If  the  difference  of  level  between 


FIGURE  12 


s 

ured  to  be  more  than  400  pounds    than    three    inches,    it   is    impossible    to 
per  square  foot.  gpring    an     arch     between     thenij     and    a 

"shelf  angle"  must  be  used  as  seen  in  Fig.  n.  In  other  words, 
an  arch  can  be  sprung  between  a  15"  beam  and  an  18"  beam  if 
the  upper  flanges  of  these  beams  are  "flush"  but  a  shelf  angle 
must  be  used  if  it  is  desired  to  spring  the  arch  between  a  15" 
beam  and  a  20"  or  a  24"  beam.  The  fifteen-inch  beam  is  considered 
as  a  standard. 


ENGINEERING  FOR  ARCHITECTS        17 

The  method  of  finding  the  weight  of  a  square  foot  of  floor 
construction,  as  given  above,  is  the  one  used  in  all  cases  where 
floor  loads  are  considered.  Sidewalk  loads  and  roof  loads  differ 
from  floor  loads  as  brick  arches  are  used  and  the  construction  must 
be  waterproof. 

In  Fig.  12  a  typical  form  of  sidewalk  construction  is  shown. 
When  a  concrete  finish  is  desired  a  cinder  concrete  fill  is  used.  The 
weight  for  a  square  foot  of  sidewalk  construction  is  obtained  as 
follows : 

3^"  cement  at  iclb  per  inch =     35lb 

2j"  concrete  fill  at  7.5lb =     19" 

i "  waterproofing •  =       5  " 

3f"  cinder  fill  at  4R> =     15" 

8"  brick  arch  at  iisib =    77" 

6olb  beam  —  6/-o'/  span =     10  " 

Total  dead  load =  161  " 

Live  load  ...'..- =  300" 

Total =  461  " 

There  are  many  different  kinds  of  sidewalk,  roof,  and  floor  con- 
struction, and,  of  course,  the  architect  must  know  the  type  that 
is  being  used  in  any  building  he  is  designing.  If  he  employs  the 
method  given  above,  he  can  easily  determine  the  weight  of  the  floor 
even  though  the  dimensions  or  the  materials  may  be  different. 

Wall  loads  are  no  harder  to  determine  than  floor  loads,  but,  as 
walls  are  cut  up  by  windows,  the  work  required  to  find  the  points 
where  the  greatest  load  is  going  to  bear  upon  the  wall  beam  is 
tedious.  If  accuracy  is  required,  it  is  necessary  to  study  every 
panel  of  the  wall.  As  the  method  of  determining  the  beam  to  carry 
a  wall,  where  the  loading  is  not  uniform,  involves  the  consideration 
of  concentrated  loads,  wall  loads  will  be  taken  up  in  the  next  chapter. 

As  problems  for  practice,  it  is  suggested  that  the  architect  deter- 
mine for  himself  whether  the  beams  in  any  framing  plan  that  he  may 
have  are  of  the  proper  size  to  carry  the  loads  found  by  him. 


CHAPTER   III 

Concentrated  Loads.  Reactions.  Points  of  maximum  bending  moment,  and  shear 
diagrams.  Method  of  finding  Af,  the  maximum  bending  moment.  Method  of 
finding  the  section  modulus  when  M  is  measured  in  foot-tons.  The  design  of 
spandrel  beams. 

^1  r^O  determine  the  size  of  beams  strong  enough  to  carry  a  series 
JL  of  concentrated  loads,  it  is  necessary  to  study  the  theory 
of  the  subject.  Our  American  ideal  of  being  practical  is  not  to  be 
condemned,  but,  unless  the  architect  grasps,  to  a  small  extent,  the 
theory  back  of  all  this  engineering  work,  he  is  apt  to  become  the  slave 
of  the  handbook. 

In  Fig.  13  a  diagram,  representing  a  simple  beam  carrying  three 
concentrated  loads,  is  shown.  The  beam  has  a  span  of  twenty  feet 
and  the  loads  are  five,  four,  and  two  tons  respectively  located  eight, 


I'-o'l 

•4TON3 

'KL. 

f 

^2 

\ 

FIGURE  13 

fifteen,  and  nineteen  feet  from  the  left  support.  In  Chapter  I,  where 
a  simple  beam  was  considered,  there  was  a  uniformly  distributed 
load  and  the  forces  exerted  at  each  support  were  equal.  In  this 
case,  however,  it  is  obvious  that  there  will  be  a  different  force  at 
-#i  from  that  at  R%.  The  loads  are  placed  on  the  right  side  of  the 
beam  and  so,  at  the  first  inspection,  it  might  be  assumed  that  the 
right  reaction  will  be  greater  than  the  left.  In  order  to  find  out  if 
this  is  true,  and,  at  the  same  time,  to  determine  the  exact  amount 
of  loading  on  each  support,  the  following  process  is  employed. 


ENGINEERING  FOR   ARCHITECTS  19 

If  R2  were  removed,  and  the  beam  were  allowed  to  swing  freely 
around  RI,  there  would  be  a  tendency  to  rotate  around  the  left  sup- 
port as  shown  in  Fig.  133.  Remembering  the  definition,  in  the  first 
article,  that  a  moment  is  the  tendency  to  produce  rotation  around 
a  point,  known  as  the  center  of  moments,  it  is  plainly  seen  that 
there  would  be  a  moment  around  RI,  and  this  will  equal  the  sum  of 
all  the  moments  caused  by  all  the  loads,  R\  becomes  the  center  of 
moments.  The  five-ton  load  will  produce  a  moment  of  40  foot-tons 
around  R\.  The  four-ton  load,  having  a  lever  arm  of  fifteen  feet, 
will  produce  a  moment  of  60  foot-tons.  A  smaller  moment  of  38 
foot-tons  will  be  caused  by  the  two  tons  located  nineteen  feet 
from  RI.  As  a  result  the  total  tendency  to  rotate  around  the  left 
support  is  given  as  follows: 

5  tons  X     8  feet=     40  foot-tons. 

4    "     x  15     "  -      60     "    " 
2    «     x  19     "  =  _38     "    " 

138      "    "  Total. 

Unless  an  upward  moment  is  used  to  counteract  this  downward 
moment  there  will  be  rotation  around  the  left  reaction.  The  only 
upward  force  that  can  possibly  produce  this  moment  is  the  right 
reaction  (R2).  This  reaction  acts  at  a  distance  of  twenty  feet  from 
RI,  and  the  upward  force  that  it  would  have  to  exert  on  the  beam 
to  produce  equilibrium  is  given  by  the  following  equation:  R%X  20 
feet  =138  foot-tons,  or  R<z  =  138-7-  20  =  6.9  tons. 

To  obtain  the  left  reaction,  the  same  method  can  be  employed, 
the  only  difference  being  that  R%  will  be  taken  as  the  center  of  mo- 
ments. In  this  case  the  results  are: 

2  tons  x     i  foot  =     2  foot-tons. 

4  "  X  5  ^et  =  20  " 
_5  "  X  12  "  =  60  « 
ii  tons  82  "  "  Total. 

82  -f-  20  =  4.1  tons  as  the  load  on  RI. 

As  a  rule  the  left  reaction  is  obtained  in  a  much  simpler  manner. 
The  sum  of  the  reactions  must  equal  the  sum  of  all  the  downward 
loads,  otherwise  there  would  be  a  tendency  to  push  the  beam  either 
up  or  down.  If  the  right  reaction  is  obtained,  the  left  one  is  deter- 
mined by  simply  subtracting  RZ  from  the  total  load.  By  arranging 
the  computation  as  shown  above,  the  total  load  is  given  by  plain 


20        ENGINEERING  FOR  ARCHITECTS 

addition.  From  the  n  tons  subtract  6.9  tons,  and  the  remainder 
is  4.1  tons.  The  second  computation  is  unnecessary  except  as  a 
check. 

Take  a  second  problem,  the  diagram  for  which  is  shown  in  Fig. 
14.  Here  we  have  a  uniform  load  as  well  as  two  concentrated  loads. 
There  need  be  no  hesitation  about  attacking  the  new  condition,  as  a 
uniform  load  is  treated  in  exactly  the  same  manner  as  a  concen- 
trated one.  First  obtain  the  total  uniform  load.  Then  find  the 
distance  from  the  center  (center  of  gravity)  of  the  load  to  the  point 
taken  as  the  center  of  moments.  The  moment  around  this  center 
is  the  product  of  the  total  load  multiplied  by  the  distance.  In  other 

words,  the  uniform  load  acts  ex- 
actly  like  a  concentrated  load 
which  has  been  placed  at  the 
center  of  gravity  of  the  dis- 
tributed weight.  The  load  per 

FIGURE  14  f°ot  ls  usually  denoted  by  small 

w,  and  the  total  weight  by  large 
W.  If  /  is  the  span,  and  w  the  weight  per  foot,  then  W  =  wl. 
If  the  architect  always  uses  the  large  W  as  a  basis  for  his  calcula- 
tion, he  will  avoid  many  disturbing  complications  that  follow  the 
use  of  the  smaller  letter.  In  the  diagram,  shown  in  Fig.  14,  R%  is 
obtained  by  the  following  calculation: 

2  tons  X    3  feet  =    6  foot-tons. 

3  "    x    7    "  =  21    "    " 
0.3  tons  x  10  =  3     "     x    8    "   =   24    "     " 

_6     "     x  14    "  ==_84    "    " 
14  tons  135    " 

T35  -*•  J5-5  =  8.71  tons  =  R^. 
14  -  8.71  =  5.29  tons  =  Ri. 

The  architect  can  check  the  calculations  by  taking  moments 
around  R2.  It  is  worth  noting  that  if  Ri  is  taken  as  the  center  of 
moments,  Rz  is  obtained.  One  of  the  most  frequent  mistakes  of  begin- 
ners is  to  give  the  value  of  the  reaction  obtained  by  the  calculations  to 
the  reaction  used  as  the  center  of  moments. 

The  fact  that  it  is  always  easy  to  check  results  makes  it  possible 
for  the  architect  to  originate  his  own  problems  and  be  absolutely 
sure  that  his  answers  are  correct.  He  can  take  spans  of  any  conven- 


ENGINEERING  FOR  ARCHITECTS        21 

ient  length  —  twelve,  sixteen,  or  twenty  feet  —  and  can  assume 
any  kind  of  loading.  If  the  reactions  are  found  in  the  proper  manner 
the  results  will  always  check. 

In  case  we  have  a  simple  beam,  unsymmetrically  loaded,  the  first 
calculations  are  made  to  find  the  reactions.  The  second  process  is 
to  find  the  maximum  bending  moment.  When  there  are  concentrated 
loads,  which  may  be  placed  in  any  position  whatever,  the  point 
where  the  maximum  bending  moment  is  going  to  occur  is  unknown. 

It  is  obvious,  that  when  there  is  a  single  concentrated  load,  as 
shown  in  Fig.  15,  the  maximum  bending  is  going  to  take  place  directly 
under  the  load.  Take  the  point  c,  at  the  load,  as  the  center  of  mo- 
ments, and  determine  the  tendency  to  produce  bending  around  this 
point.  The  right  reaction  is  8  tons  —  check  this  —  and  the  distance 

from  c  is  5  feet.     So  the  bend-  , 

i  j    i  h to'-o 4* •5'-<?  *  -vt 

ing  around    c  is  caused    by  a  T/2r 

moment  of  8   tons  X  5  feet  =    f^TT^  *-c          •sPtsr 

40  foot-tons.    The  left  reaction    !*—  —15-0'—  -Aj 

(R)i  causes  a  bending  moment  FIGURE  15 

of  4  tons  x  10  feet  =  40  foot- 
tons,  which  equals  that  produced   by  R^.     This  is  correct,  for,   if 
one  moment  were  greater  than  the  other,  rotation  would  take  place 
around  c  and  the  beam  would  not  be  in  a  condition  of  equilibrium. 

The  first  impulse  of  a  beginner  is  to  use  the  1 2-ton  load  to  find  the 
bending  at  c.  It  seems  plain  that  this  load  causes  bending,  and,  as 
it  creates  the  loads  on  the  reactions,  it  does.  But  the  1 2-ton  load 
acts  through  the  point  c.  There  is  no  lever  arm,  and,  therefore,  it 
produces  no  direct  bending. 

.The  fact  that  the  maximum  bending  occurs  at  the  point  c  is  so 
plain  that  it  needs  no  further  explanation.  When,  however,  there 
are  several  loads  as  in  Figs.  13  and  14,  it  requires  some  calculation 
to  find  the  exact  spot  where  the  greatest  moment  is  going  to  occur. 

The  next  statement  must  be  taken  by  the  architect  without 
proof.  That  is:  the  maximum  bending  occurs  where  zero  shear 
exists.  This  means  that  where  there  is  no  tendency  for  the  beam 
to  fail  by  shearing  it  is  most  apt  to  fail  by  bending.  To  prove  this 
it  is  necessary  to  resort  to  calculus,  but  all  that  the  architect  need 
realize  is,  that  to  find  the  point  of  greatest  bending,  it  is  necessary 
to  find  the  point  of  no  shear. 

Fig.  3,  in  Chapter  I,  shows  a  beam,  uniformly  loaded,  failing  by 
shearing  at  the  supports.  In  any  case  the  greatest  shear  is  found 


22 


ENGINEERING  FOR  ARCHITECTS 


at  the  reactions  and  in  the  case  of  the  uniformly  distributed  load 
there  can  be  no  tendency  to  shear  at  the  center  of  the  beam,  as 
the  upward  reactions  (Rt  =  R%  =  \W)  would  be  counteracted  by  the 
downward  weight  of  one-half  of  the  wall  at  the  right  or  left  of 


FIGURE  16 

the  center.  At  this  point  the  downward  force  becomes  equal  to 
the  upward  force  and  the  shear  becomes  zero.  The  diagram  that 
expresses  this  is  shown  in  Fig.  16.  Lay  off  a  distance,  upward,  at 
RI  equal  to  the  left  reaction,  and  lay  off  at  R%  a  distance  downward, 
equal  to  the  right  reaction.  At  the  center  there  is  no  shear.  Con- 
nect the  points  with  a  straight  line  and  the  shear  diagram  is  drawn. 
The  shear  at  the  left  support  equals  RI  and  steadily  decreases  as  the 
distance  increases  away  from  the  reactions,  until  at  the  center  the 
shear  is  zero.  From  there  on  the  shear  increases  until  it  reaches 
the  right  support  where  it  equals  R%.  The  line  representing  the  shear 
caused  by  a  uniform  load  is  always  a  sloping  line  and  the  total  drop 
is  equal  to  the  total  load. 

Concentrated  loads  have  a.  different  diagram.     In  Fig.   14,  if  a 
weak  spot  should  occur  at  any  point  between  the  left  support  and 


FIGURE  17 

the  first  load,  the  upward  reaction  (Ri)  would  shear  off  the  beam  at 
that  point.  The  shear  will  remain  the  same  for  all  points  between 
the  reaction  and  the  first  load.  This  is  shown  in  the  shear  diagram 
(Fig.  17).  At  the  point  where  the  first  load  is  located,  the  shear 
caused  by  R]_  will  be  offset  by  the  downward  load  of  five  tons.  This 


ENGINEERING  FOR  ARCHITECTS 


will  cause  a  minus  shear  of  4.1  —  5.0=  —  0.9  tons  shear.  There 
would  be  no  change  in  shear  until  the  next  load  is  reached  where 
the  minus  shear  of  —  0.9  —  4.0  =  —  4.9  tons  will  occur.  The  two- 
ton  load  will  cause  a  minus  shear  of  -  4.9  -  2.0  =  —  6.9  tons,  which 
will  remain  the  same  for  all  points  between  the  last  load  and  the 
right  reaction.  When  R2  is  reached  the  upward  force  of  6.9  just 
counteracts  the  downward  (minus)  shear. 

In  Fig.  1 8  a  shear  diagram  is  shown  for  the  loading  given  in 
Fig.  14.  The  only  difference  between  this  diagram  and  that  in  Fig. 
17,  is  that  the  uniform  load  gives  a  sloping  line,  instead  of  a  "step" 
as  a  concentrated  load  would.  The  total  drop  of  the  sloping  line 
is  equal  to  the  total  uniform  load.  The  distance  "a"  plus  the 
distance  "b"  should  equal  3  tons.  The  shear  diagram  gives  graphi- 
cal proof  that  the  sum  of  the  reactions  must  equal  the  sum  of  the 
loads. 

For  the  beam  shown  in  Fig.  13,  the  loads  will  give  a  maximum 
bending  moment  at  the  point  where  the  five-ton  load  is  located,  as 


FIGURE  18 


the  shear  passes  through  zero  at  that  point.  This  moment  must 
equal  4.1  tons  x  8  feet  =32  foot-tons.  To  prove  that  this  is  the 
maximum,  take  the  moments  at  the  four-ton  load.  (4.1  x  15)  - 
(5X7)  =  26.5  foot-tons.  The  moment  under  the  small  load  —  two 
tons  — will  be  (4-1  X  19)  -  (5  X  u)-  (4x4)  =  77.9-  (55+ 16)  = 
6.9  foot-tons.  This  result  checks  with  the  fact  that  the  moment 
caused  by  R2  around  the  two-ton  load  is  6.9  tons  X  i  foot  =  6.9  foot- 
tons.  Once  the  maximum  bending  moment  is  determined,  the 


24 


ENGINEERING  FOR  ARCHITECTS 


section  modulus  is  found  in  the  following  manner:  M  =  32  foot-tons, 
or,  32  X  12  inch-tons.  S,  taken  in  tons,  equals  8  tons.  So,  32  x  12 
=  8  X  I/c.  I/c  =  32  X  12/8  =  48.  A  fifteen-inch  I-beam,  weigh- 
ing forty-two  pounds  per  foot,  will  be  strong  enough  to  carry  this 
load. 

If  M  is  measured  in  foot-tons,  to  find  I/c  simply  multiply  M  by 
3/2.  To  prove  this  substitute  in  the  formula  M  =S  X  I/c.  M  X  12 
=  8  xl/c  or,  M  X  12  -5-  8  =  I/c.  So  M  X  3/2  =  I/c.  This  method 
can  be  used  only  when  M  is  measured  in  foot-tons. 

For  the  beam  in  Fig.  14  the  maximum  bending  is  found  at 
the  concentrated  load  of  three  tons.  M,  in  this  case,  equals 


FIGURE  19 

5.29  x  7  -  (2  x  4)  -  (0.3  x  4  X  2)  =  37.03  -  8  -  2.4  =  26.63  foot-tons. 
Taking  moments  at  the  right  as  a  check,  the  results  are,  8.71  X  8.5 
-  (6  x  7)  -  (0.3  x  6  x  3)  =  26.63  foot-tons.  The  section  modulus 
is  26.63  X  3/2  =  39-94?  and  a  twelve-inch,  forty-pound  I-beam  will 
be  required  for  these  loads. 

The  processes,  given  above,  can  be  re-stated  briefly,  as  follows: 
First,  determine  the  reactions.  Second,  draw  the  shear  diagram 
and  find  the  point  of  no  shear.  Third,  determine  the  bending  mo- 
ment at  this  point.  Fourth,  find  the  section  modulus  and  the  size 
of  beam  required. 

Now  to  reduce  all  the  theoretical  discussion  to  practical  con- 
siderations, consider  the  wall  load,  shown  in  Fig.  19.  The  wall  is 
one  foot  thick  and  is  pierced  by  windows,  which  are  5'-8"  wide  by 
9/-o'/  high.  Below  the  windows  and  running  the  entire  length  of  the 


ENGINEERING  FOR  ARCHITECTS 


wall  is  a  stone  course  which  is  two  feet  thick.  The  sill  of  the  win- 
dows is  two  feet  above  the  flange  of  the  beam.  The  steel  columns 
are  2o'-o"  apart  and  are  located  so  that  the  loading  on  the  wall  beams 
will  be  symmetrical.  The  distance  between  the  upper  flanges  of  the 
wall  beams  will  be  12'-^".  This  gives  a  wall  panel  12'-^"  X  2o'-o". 
The  weight  of  the  brick  over  the  windows  is  carried  down  to  the 
beam  by  the  piers  and  the  mullion.  These  brick  weights  will  be 
considered  as  concentrated  loads  and  the  stone  will  be  taken  as  dis- 
tributing its  weight  uniformly  over  the  entire  beam.  The  loading  of 


FIGURE  20 


the  beam  is  shown  in  Fig.  20.     To  get  these  loads  the  following  pro- 
cess is  employed: 

The  distance  from  the  center  line  of  the  columns  to  the  center 
line  of  the  windows  is  6^4 "  and  the  area  of  brick  included  between 
these  lines  is  (io'-4"  X  6'-4")  minus  one-half  the  window  area  which 
is  (Q'-O"  X  2/-io").  This  gives  65.4  square  feet  minus  25.5  square 
feet,  which  equals  approximately  40  square  feet.  As  the  wall  is  one 
foot  thick,  the  weight  of  this  area  is  40  x  i  X  120=  4,800  pounds. 
This  is  the  load  brought  down  by  each  pier.  The  load  brought 
down  by  the  mullion  is  obtained  by  finding  the  weight  of  the  brick 
area  between  the  center  lines  of  the  two  windows.  (7'-4"  X  io'-4") 
-  (5 '-8"  X  9'-o")  =  24.77  square  feet.  24.7  x  I  X  120  =  approxi- 
mately 3,000  pounds.  The  uniform  load  of  the  stone,  which  is  two 
feet  high,  two  feet  thick,  and  twenty  feet  long,  is  80  x  160  =  12,800 
pounds.  As  the  loads  are  symmetrically  distributed,  each  reaction 
will  be  equal  to  one-half  of  the  total  load,  or  4,800+  3,000+  12,800 
+  4,800  =  25,400/2  =  12,700  pounds. 


J2700 


26  ENGINEERING  FOR   ARCHITECTS 

The  maximum  bending  moment  will  be  in  the  center  of  the  beam 

and  will  be  equal  to  the  sum  of  all  the  moments  taken  at  the  left 

of  this  point.     12,700  X  120"  -  (4,800  X  99"  +  12,800/2  X  60 "  +  1,500 

X  5")  =  1,524,000     inch-pounds-  866,700     inch-pounds  =  657,300 

inch-pounds. 

To  find  the  section  modulus  of  the  beam,  divide  the  bending 
moment  by  the  stress  per  square  inch  allowed  for  steel  -  16,000 
pounds.  657,300  -r-  16,000  =  41  approximately.  A  twelve-inch 
I-beam,  weighing  forty  pounds  per  foot  will  be  strong  enough  to  carry 
the  loads.  From  the  above  calculations,  it  can  be  seen  that  it  is 
much  simpler  to  use  foot-tons  rather  than  inch-pounds. 

There  are  all  kinds  of  conditions  where  concentrated  loads  occur. 
When  there  is  framing  around  elevator  shafts,  and  other  cases  where 
beams  frame  into  girders  in  a  manner  that  gives  unsymmetrical  load- 
ing, it  is  necessary  to  use  the  method's  employed  in  this  article. 
The  method  of  drawing  bending  moment  diagrams  and  the  considera- 
tions involved  in  designing  a  built-up  girder  will  be  taken  up  in  the 
next  chapter.  For  the  present  it  would  be  well  for  the  architect  to 
become  thoroughly  acquainted  with  shear  diagrams  and  points  of 
maximum  bending  moments. 


CHAPTER   IV 

"Built  up"  Girders.  Depth  of  girder.  Thickness  of  web  plate.  Maximum  bending 
moment.  Area  of  flanges.  Length  of  cover  plates  and  bending  moment  dia- 
gram. Stiffeners. 

THE  standard  beams,  rolled  by  the  Steel  Companies,  are  strong 
enough  to  be  used  under  ordinary  conditions,  but  when  the 
spans  are  very  great  and  the  loads  unusually  heavy,  these  beams  can- 
not be  made  use  of.  Special  girder  beams  are  now  being  rolled  which 
withstand  much  more  loading  than  the  ordinary  I-beam,  but  there  are 
many  conditions  which  make  it  necessary  to  use  riveted  girders. 

Riveted,  or  "built  up"  girders,  are  made  of  plates  and  angles, 
fabricated  in  such  a  manner  as  to  give  a  section  very  similar  to  that 
of  an  I-beam.  The  diagram,  shown  in  Fig.  22,  <- COVE:^ 
shows  the  parts  of  a  single  web-girder.  The 
flange  is  made  of  cover  plates  and  angles  and 
the  web  is  made  of  a  single  rolled  plate  usually 
from  three-eighths  of  an  inch  to  one  inch  thick. 
Girders  are  made  with  two  and  sometimes  three 
webs,  but  as  these  are  difficult  to  fabricate,  the 
singleweb-girder  should  be  used  wherever  possible. 

There  are  no  two  engineers  who  use  exactly  the 
same  method  in  designing  riveted  girders,  but  the 
results  given  by  the  method  employed  in  the  fol- 
lowing problems  have  always  been  satisfactory  and  can  be  used  safely. 

Given  a  girder  with  a  clear  span  of  36  feet,  a  uniformly  distrib- 
uted load  of  80  tons,  and  one  concentrated  load  of  60  tons  and  another 
of  50  tons,  respectively  located  12  f3et  and  26  feet  from  the  left  sup- 
port; the  first  step  is  the  same  as  in  the  case  of  a  simple  beam,  namely, 
to  determine  the  reactions.  The  method  is  the  same  as  used  in 
Chapter  III. 

60  tons  X  12  feet  =     720  foot-tons. 
80     "    x  18    "    =  1440 
jo     "    x  26    "    =  1300 
190  tons  3460  foot  tons. 

3460  foot  tons  -=-36  feet  =  96  tons  =  R*. 
190  tons  —  96  tons  =  94  tons  =  RI. 


28 


ENGINEERING  FOR  ARCHITECTS 


Jjc 


-i 

< 


u= 


TS 


FIGURE  23 


The  maximum  shear  is  equal  to  the  maximum  reaction  and  is 
therefore  96  tons. 

The  depth  of  the  girder  is  considered  next.  Usually  the  textbooks 
give  the  assumed  depth  of  built-up  girders  as  a  certain  ratio  of  the 
span.  This  ratio  is  usually  assumed  as  one-ninth  or  one-tenth. 

For  a  girder,  36  feet  long,  the  depth  would  be  about  3  feet  6  inches, 
or,  42  inches.  This  depth  is  the  distance  between  the  backs  of  the 
flange  angles. 

The  difficulty  arising  from  the  use  of  the  above  ratio  is  that  no 
account  is  taken  of  the  loading  on  the  girder.  A  formula,  based  on 

pure  assumptions,  but  which  gives  satis- 
factory results,  is  often  used.  The  formula 
is,  d  =  zF/R,  in  which  V  is  the  maximum 
shear  and  R  is  the  shearing  value  of  the 
rivets  used  in  fabricating  the  girder.  To 
use  this  formula  it  is  necessary  to  assume 
the  size  of  the  rivets,  but  as  a  rule,  three- 
quarter  or  seven-eighth  inch  rivets  are 
used  in  all  such  work.  Providing  we 
assume  the  smaller  diameter,  we  must  find 
X  •  — i, r-  the  value  of  three-quarter  inch  rivets  in 

^—s.  VA  :_i,L 

double  shear. 

When  two  angles  are  riveted  to  a  plate, 
as  shown  in  Fig.  23,  then  the  rivets  are 
said  to  be  in  double  shear.  When  one 
angle  is  used  (Fig.  233),  then  the  rivets 
are  in  single  shear. 

Under  the  heading  of  "  Shearing  Values 

for  Rivets"  in  handbooks,  these  values  are  given  for  unit  shearing 
stresses  from  6,000  pounds  per  square  inch  to  10,000  pounds  per 
square  inch.  In  the  1909  edition  of  the  Cambria  Steel  Company's 
book,  page  316,  the  shearing  value  of  a  three-quarter  inch  rivet, 
having  a  unit  shearing  strength  of  10,000  pounds  per  square  inch, 
is  given  in  double  shear  as  8,836  pounds,  or,  roughly,  as  4.4  tons. 

The  vertical  shear  is  96  tons  so,  as  d  =  2  V/R,  d  =  2  x  96/4.4  =  44 
inches. 

This  " d"  is  known  as  the  effective  depth,  and  is  not  the  same  as  the 
one  given  by  the  use  of  the  ratio.  Fig.  24  shows  a  section  of  the 
girder,  and  between  the  rows  of  rivets  in  the  flange  angles,  center 
lines  are  drawn.  The  distance  between  these  center  lines  is  ((d." 


ENGINEERING  FOR  ARCHITECTS 


29 


To  find  the  location  of  the  center  lines,  it  is  necessary  to  find 
the  distances  from  the  back  of  each  angle  to  the  centers  of  the  rivet 
holes.  These  distances  are  usually  known  as  the  "gauge."  Fig.  23 
gives  the  gauge  for  a  6-inch  leg  and  the  distance  from  the  back  of  the 
angle  to  the  center  of  the  rows  of  rivets  is  i\"  +  i|"  =  3!"  as  shown 
in  Fig.  24.  3 f  "  -f  44 "  +  3f "  =  51}"  or  roughly  52  inches  which  equals 
the  distance  from  the  back  to  the  back  of  the  flange  angles.  This 


F  L/^MGE^  ^  N  G  L.E1 


FIGURE  24 

is  taken  as  the  depth  of  the  girder,  and  the  use  of  the  formula  always 
gives  a  greater  depth  than  that  obtained  by  the  ratio. 

In  construction,  the  depth  is  often  limited  by  conditions  such 
as  the  thickness  of  floors  or  the  available  head  room,  but  where  it  is 
possible  to  use  a  deep  girder,  the  depth  obtained  by  the  above  formula 
gives  good  results,  as  the  deeper  the  girder  the  smaller  the  flange 
area.  In  this  case  a  depth  of  4  feet,  or  48  inches,  will  be  used. 

The  next  step  is  the  determination  of  the  thickness  of  the  web. 
This  can  be  found  directly  from  the  table  giving  the  shearing  value 
of  rivets,  as  this  table  also  gives  the  bearing  values  of  riveted  plates. 
The  shearing  value  of  a  three-quarter  inch  rivet,  in  double  shear  is 
8,836  pounds.  Following  to  the  right,  it  is  found  that  a  J-inch  plate 
has  a  bearing  value  of  3,750  pounds,  a  9/i6-inch  plate  has  a  bearing 
value  of  8,438  pounds,  and  a  f-inch  plate  has  a  value  of  9,375  pounds. 
Under  this  last  value  a  black  line  is  drawn,  showing  that  9,375  pounds 


30        ENGINEERING  FOR  ARCHITECTS 

is  about  the  same  as  8,836  pounds,  the  shearing  value  of  the  rivet. 
A  9/i6-inch  plate  would  give  a  value  a  little  less  than  that  of  the 
rivet.  If  the  web  plate  should  be  f-inch  thick,  there  would  be  no 
more  tendency  for  the  plate  to  fail  by  bearing  than  for  the  rivet  to 
fail  by  shearing. 

The  above  process  gives  the  thickness  of  the  web  plate,  but  this 
must  be  checked  to  determine  whether  the  web  will  be  strong  enough 
to  resist  the  shear.  Flanges  resist  bending  and  webs  resist  shearing. 
The  maximum  shear  is  96  tons.  The  area  of  the  plate  is  48"  x  f "  =  30 
square  inches.  Assuming  a  shearing  value  of  4.5  tons  per  square 
inch  the  value  of  the  plate  is  given  by  30  x  4.5  =  135  tons,  which  is 
considerably  greater  than  necessary  to  withstand  the  shear. 

So  far  we  have  determined  the  depth 
of  the  girder  and  the  thickness  of  the  web 
plate.  The  next  step  is  the  determination 
of  the  flange  members.  The  formula  used 
in  this  case  is  M  =  SAD,  a  fairly  easy  one 
to  remember.  M  equals  the  maximum 
bending  moment,  S  equals  the  safe  tensil 
or  compressive  stress  of  steel,  A  equals 
the  area  of  the  flange,  and  D  equals  the 
^  depth.  To  find  A  we  must  first  find  M. 

The   shear  diagram,   shown   in    Fig.    25, 

gives  the  point  of  maximum  bending  moment  as  15.3  feet  from 
RI.  The  method  employed  in  drawing  this  shear  diagram  was 
explained  in  Chapter  III. 

To  find  the  maximum  bending  moment  at  this  point  first  find  the 
upward  moment  caused  by  the  reaction.  94  tons  x  15.3  feet  =  1438.2 
foot-tons.  The  downward  moments  caused  by  the  loading  are  2.22 

X  15.3  -   -X  =  259.84   foot-tons,    and   60  X  3.3  =  198    foot-tons,  or 

a  total  downward  moment  of  457.84  foot-tons.  The  total  maxi- 
mum bending  moment  at  this  point  then  is  1438.2  —  457.8  =980 
foot-tons  approximately. 

Now,  as  M  =  SAD  and  M  equals  980  foot-tons,  S  in  the  case 
of  riveted  steel  equals  7.0  tons,1  and  D  equals  48  inches,  A  can  be 

r        ,        .      M          980  .     , 

round.    A  =—  —  =  35  square  inches. 

SD      4  X  7.0 

1   The  New  York  Building  Department  now  will   allow   a   safe  working  stress  of 
8  tons  per  square  inch  on  net  area. 


ENGINEERING  FOR  ARCHITECTS 


This  area  is  made  up  of  two  angles  and  cover  plates,  and  to 
determine  the  size  of  these  members  some  experimenting  is  required. 
Good  practice  requires  that  all  members  should  have  about  the 
same  thickness.  It  would  be  bad  design  to  have  an  angle,  I  inch 
thick,  riveted  to  a  plate  f  inch  thick.  As  we  have  already  assumed 
that  one  leg  of  each  flange  angle  is  6  inches  long  we  can  assume  that 
the  length  of  the  other  leg  is  also  6  inches,  and  the  thickness  is  the 
same  as  that  of  the  web  plate  or  f  inch.  So  the  flange  angles  will 
be  6  x  6  x  f  inch. 

Under  the  heading  of  "Properties  of  Standard  Angles-Equal 
Legs"  the  area  of  a  6x  6x  f  inch  angle  is  given  as  7.11  square 
inches.  Two  angles  would  have  an  area  of  14.22  square  inches. 
The  total  area  of  the  flange  being  35  square  inches,  the  area  left  to 
be  made  up  by  the  cover  plates  will  be  35  —  14.22  =  20.78  square 
inches.  The  width  of  the  cover  plates  can  be  taken  as  14  inches, 
and  20.78  square  inches  divided  by  14  inches  will  give  the  total 
thickness  of  cover  plates  as  ij  inches.  The  thickness  can  be  made 
up  of  three  J-inch  plates. 

The  bottom  flange  is  in  tension  and  a  portion  of  the  material  is 
lost  because  of  the  rivet  holes.  Fig.  26  shows  a  portion  of  the 
flange  and  if  the  plates  should  fail  by 
tension  —  pull  a  part  —  the  failure  would 
occur  on  either  line  BB  or  AA.  It  is 
practically  certain  that  failure  could  not 
occur  on  line  CC.  There  would  be  two 
rivet  holes  in  the  fractured  section.  For 
f-inch  rivets,  holes  are  punched  f  inch 
in  diameter.  Each  hole  pierces  three 
J-inch  plates  as  well  as  an  angle  f  inch 
thick.  So  the  area  lost  by  each  rivet  hole  will  be  (i  J  inches  +  f  inch) 
X  1  inch  =  1.86  square  inches  and  two  holes  will  cause  a  loss  of 
3.72  square  inches. 

In  the  case  of  the  upper  flange  20.78  square  inches  had  to  be  made 
up  by  plates,  and  as  3.72  square  inches  are  lost  by  rivet  holes  in  the 
bottom  flange,  20.78  +  3.72  =  24.50  square  inches  must  be  made  up. 
The  total  thickness  is  given  by  24.50  -=-  14  =  if  inch,  and  the  flange 
will  have  one  cover  plate  \  inch  thick  and  two  f  inch  thick.  To 
facilitate  the  fabrication  of  the  girder  both  flanges  are  made  alike 
and  the  plates  are  cut  the  same  length. 

To  determine  the  length  of  the  cover  plates  the  bending  moment 


U 

FIGURE  26 


ENGINEERING  FOR  ARCHITECTS 


diagram  is  made  use  of.  Nothing  has  been  said  about  the  method 
of  drawing  bending  moment  diagrams,  as  they  are  seldom  used 
except  in  such  problems  as  this.  Under  the  heading  of  "Bending 
Moments  and  Deflections  of  Beams  of  Uniform  Sections"  several 
types  of  diagrams  are  shown  in  the  handbooks.  When  a  beam  is 
uniformly  loaded  the  bending  moment  diagram  takes  the  form  of  a 
parabola. 

The  method  of  drawing  a  parabola  is  simple,  and  as  it  is  necessary 
to  know  this  in  order  to  draw  the  final  diagram  for  the  girder  the 
process  is  worth  mastering.  The  maximum  bending  moment  for 
a  uniform  load  is  given  by  the  formula  M  =  •§•  Wl.  In  case  there  is 
a  uniform  load  of  80  tons,  on  a  girder  36  feet  long,  the  maximum 
bending  moment  is  \  X  80  X  36  =  360  =  foot-tons. 

In  Fig.  27  the  line  ab  is  laid  off  to  represent  36  feet,  for  the  pur- 
poses of  demonstration,  say  at  a  scale  of  f-inch  equals  I  foot.  This 


O       10    .3876      777 


/    -- 1 

£ 

3 

4 

51 


FIGURE  27 

is  the  base  of  the  parabola  and  is  equal  to  the  length  of  the  girder. 
The  line  cm  is  laid  off  to  represent  the  maximum  bending  moment. 
If  an  inch  is  considered  as  representing  200  foot-tons,  1.8  inches 
will  represent  360  foot-tons.  Draw  ao,  oo' ',  and  o'b  and  divide  oa 
into  six  parts.  Also  divide  om  into  an  equal  number  of  parts,  in  this 
case  six.  Starting  from  o  on  oa  mark  the  points  I,  2,  3,  4  and  5 
and  from  m  on  mo  mark  the  points  6,  7,  8,  9  and  10.  From  m  draw 
radiating  lines  to  I,  2,  3,  4  and  5,  and  from  6  draw  a  vertical  line 
to  mi,  from  7  to  m2,  and  so  until  from  10  a  vertical  line  intersects 
m5-  These  points  of  intersection  are  points  on  the  parabola  and  by 
joining  them  the  bending  moment  diagram  for  a  uniform  load  of 
80  tons  on  the  girder  is  completed. 

Again  referring  to  the  handbook,  the  bending  moment  diagram 
for  a  single  concentrated  load  is  seen  to  be  a  triangle,  the  base  being 


ENGINEERING  FOR  ARCHITECTS 


33 


equal  to  the  span  and  the  altitude  equal  to  the  maximum  bending 
moment.  To  determine  the  value  of  this  maximum  bending  mo- 
ment, due  to  the  single  concentrated  load,  the  following  method  is 
employed. 

In  Fig.  28,  a  girder,  with  a  span  denoted  by  "/,"  is  loaded  with  a 
single  load   "W"   at  a   distance   "a"  from    Rl  and   \'b"  from  R2. 


FIGURE  28 

Then  the  downward  moment  around  Ri  equals  W  x  a  and  this  equals 
the  upward  moment  of  R2  x  /,  or,  Wa  =  R2l,  R2  then  must  equal  — — 


In   the  same  manner  R1 


-  . 


The  bending  moment  at  c  equals 


Wb  W 

Ri  X  a  or  —  x  a  =  —  X  ab.     This  checks  with  the  moment  of  R2 

Wa  W 

around  c,  or,  -  x  b  =  —  Xab. 

In  the  case  of  a  girder  having  a  span  of  36  feet  and  a  load  of  60 

tons  located  12  feet  from  RI,  the  maximum  moment  caused  by  the 

load    will   be  f£  x  12x24=480  foot-tons,   and   for  a   load   of   50 

tons,  10  feet  from  R2  the  maximum  moment  will  be  ft  X  10  X  26 

=  361.1  foot-tons. 


34        ENGINEERING  FOR  ARCHITECTS 

In  Fig.  29  all  the  diagrams  described  above  are  shown.  The 
two  triangles  represent  the  bending  moments  caused  by  each  con- 
centrated load.  The  parabola  represents  the  bending  due  to  the 
uniformly  distributed  load.  The  line  ABCDE  represents  the  sum 
of  all  the  moments  due  to  all  the  loads  and  is  therefore  the  actual 
bending  moment  diagram.  The  point  B  is  found  by  the  addition  of 
OX,  OT,  and  6>Z,  or,  in  other  words,  OB,  =  OX+OT  +  OZ.  In 
the  same  manner  all  other  points  on  the  diagram  are  found.  It  will 


/A  Q 

FIGURE  29 

be  found  that  the  maximum  bending  occurs  at  C  which  is  15.3  feet 
from  A. 

Now  the  question  naturally  arises,  what  is  the  use  of  all  this? 

Disregarding  all  the  more  or  less  complicated  formulas  employed 
to  prove  that  the  following  processes  are  correct  the  next  step  will 
be  to  find  the  length  of  the  cover  plates. 

In   Fig.   30,  the  bending  moment  diagram,  already  determined, 
is  shown.     The  area  of  each  flange  is  made  of  two  angles  —  6  inch  x 
6  inch  X  f  inch  —  having  an  area  of  14.22  square  inches,  one  plate  — 
14  inch  X  J  inch  —  having  an  area  of  7.0  square  inches,  and  two 


ENGINEERING  FOR  ARCHITECTS 


35 


inch, — having   together    an    area    of 
total   flange   area   of  38.72    square 


plates  —  each    14    inches  x 
17.5    square   inches,   making   a 
inches. 

Draw  through  the  point  e,  which  is  located  at  the  point  of  maxi- 
mum bending  moment,  a  horizontal  line  dd.  Between  RR  and  dd 
lay  off  a  line  AB,  at  some  convenient  scale,  having  a  length  of  38.72 
units.  On  this  line  lay  off  AX,  having  a  length  of  14.22  —  the  area 


FIGURE  30 

of  the  angles, —  XT  having  a  length  of  8.  75 —  the  area  of  one 
f-inch  plate,  7~Z,  the  same  length,  and  ZB  equal  to  7.0  units  —  the 
area  of  the  ^-inch  plate.  Through  X,  T  and  Z  draw  horizontal 
lines.  The  line  through  X  pierces  the  diagram  at  a  and  a1 ',  and 
the  length  aa' ,  is  the  theoretical  length  of  the  f-inch  plate.  The 
length  W  is  the  theoretical  length  of  the  next  f-inch  plate,  and  ccr 
is  the  length  of  the  |-inch  plate.  The  angles  of  course  run  the  full 
length  of  the  girder.  In  actual  fabrication  the  first  f-inch  plate 
is  made  to  cover  the  total  length  of  the  girder.  The  bottom  plate 


ENGINEERING  FOR  ARCHITECTS 


is  made  I  foot  6  inches  longer  on  each  end  than  is  theoretically  re- 
quired to  allow  for  riveting  and  the  f-inch  plate  is  made  at  least  2 
feet  longer  on  each  end. 

The  reason  for  giving  less  projection  for  the  bottom  plate  beyond 
that  theoretically  required,  is  because  at  one  point  only  is  this  plate 
stressed  to  its  limit,  this  point  being  where  the  shear 
is  zero.     At  all  other  points  in  the  plate  there  is  a 
little  more  material  than  necessary. 

The  only  other  step  to  be  taken  in  the  design  of  a 
riveted  girder,  that  an  architect  needs  to  know  about, 
is  the  design  of  the  stifFeners.  At  the  points  of 
support,  and  usually  under  each  concentrated  load, 
stifFeners  are  used.  Where  a  heavy  uniform  load  is 
carried  by  a  girder  stifFeners  are  riveted  in  such  posi- 
tions that  the  distance  between  them  is  no  greater 
than  the  depth  of  the  girder. 

In  Fig.  31  the  diagram  shows  how  buckling  may 
occur  in  the  web  of  the  girder.  The  web  plate  being 
only  f  inch  thick  might  easily  bend  as  shown,  pro- 
vided it  is  not  braced.  For  the  purposes  of  bracing 
angles  are  employed. 

There  are  many  formulas  now  being  used  to  deter- 
mine the  size  of  these  angles  but  the  simple  formula 
P  =  SA  is,  in  the  opinion  of  the  author,  safe  enough. 
P  equals  the  load,  S  equals  the  safe  compressive 
stress  of  riveted  steel,  and  A  equals  the  total  area  of  the 
angles. 

In  the  case  of  the  girder  P  is  the  maximum  shear  —  96  tons  — 
and  S  is  7.5  tons.  A  has  to  be  determined.  96  =  7.5  X  A,  or  A  = 
|~J  =  12.6  square  inches.  The  outstanding  leg  of  the  angle  can 
be  no  more  than  5  inches  or  it  would  overlap  the  flange  angle.  A 
single  5  inch  X  3i  inch  X  f  inch  angle  will  have  an  area  of  6.68  square 
inches.  Two  angles  will  have  about  the  area  required. 

Another  method  of  determining  the  stifFeners  is  to  consider  the 
area  of  the  outstanding  leg  of  the  stifFener  angle  only.  The  leg 
will  be  5  inches  long  and  S  in  this  case  is  20,000  pounds  per  square 
inch,  or  10  tons.  The  thickness  of  the  angle  is  found  as  follows. 
96  =  2  X  5  X  10  X  t  in  which  t  =  the  thickness. 


FIGURE  31 


9  6 


=  approximately 


ENGINEERING  FOR  ARCHITECTS  37 

These  stiffeners  are  riveted  over  the  flange  angles,  and  filler  plates 
f  inch  thick  are  riveted  between  the  web  plate  and  the  stiffeners. 
The  spacing  of  rivets  is  usually  determined  by  the  conditions 
of  fabrication.  Any  good  firm  of  steel  contractors  will  supply 
more  than  the  necessary  number  of  rivets  and  the  architect  will  have 
no  reason  to  bother  about  this  consideration. 


CHAPTER  V 

Columns.  Formulas.  Assumed  section  and  safe  stress  per  square  inch.  Dimen- 
sions. Moments  of  inertia.  Radii  of  gyration.  Actual  safe  working  stress. 
Supporting  value  of  section.  Checks. 

THE  design  of  columns  involves  some  new  considerations,  - 
such  as  the  determination  of  the  radius  of  gyration  — which 
seem  complicated  and  confusing  at  first  glance,  but  which  really  are 
simple  if  attacked  intelligently.  In  the  first  article  the  value  of 
S  —  the  safe  working  stress  for  steel  —  was  given  as  sixteen  thou- 
sand pounds  per  square  inch.  This  value  was  obtained  by  crushing 
cylinders  of  steel  in  testing  machines,  and  finding  the  force  necessary 
to  fracture  the  metal.  The  average  crushing  value  of  steel  has  been 
found  to  be  64,000  pounds  per  square  inch,  and,  as  a  factor  of  safety 
of  four  is  always  used  with  this  metal,  the  safe  stress  becomes  64,000 
-f-  4  =  16,000  pounds  per  square  inch. 

If  columns  of  steel  were  short  in  proportion  to  their  sectional 
area  they  would  act  in  the  same  manner  as  the  blocks  of  steel  crushed 
in  the  testing  machines,  but  this  is  never  the  case.  Not  only  does  a 
column  tend  to  fail  by  crushing,  but  also  by  bending.  No  one 
really  knows  just  how  the  combination  of  crushing  and  bending 
stresses  affects  the  material  in  a  column.  Formulas  have  been  de- 
rived in  which  every  theoretical  consideration  seems  to  have  been 
accounted  for,  but  these  formulas  have  all  had  to  be  modified  to 
agree  with  actual  results  obtained  from  experiments  in  which  columns 
were  subjected  to  crushing  forces. 

The  first  formula,  of  any  kind,  given  in  nearly  every  textbook 
on  engineering  is  P  =  SA  in  which  P  equals  the  compressive  load,  S 
equals  the  safe  unit  stress  and  A  equals  the  area.  If  P  were  to 
be  taken  as  300  tons,  and  S  as  8  tons  per  square  inch,  then  A  must 
equal  — g-—  =  37.5  square  inches  as  the  area  of  a  short  block  of  steel 
supporting  a  load  of  300  tons.  If  the  block  of  steel  were  not  short, 
but  had  a  height  comparatively  great  in  relation  to  its  sectional 
area,  then  it  is  obvious  that  S  would  become  less,  and  in  order  to 
support  the  300  tons,  the  area  would  have  to  be  increased. 

The  general  method  of  attacking  a  problem  pertaining  to  the 
design  of  columns  is  to  assume  that  a  smaller  value  of  S  must  be 


ENGINEERING  FOR  ARCHITECTS        39 

taken.  Once  S  is  established  the  load  is  divided  by  it,  and  the 
cross-section  area  of  the  column  is  obtained.  The  real  problem 
involved  in  column  design  is  the  determination  of  S. 

The  handbook  published  by  the  Cambria  Steel  Company  gives 
a  formula  for  S  derived  by  Gordon.  This  formula  gives  the  ulti- 
mate breaking  strength  per  square  inch  of  a  column  section  as 

50,000 

S=   "         (i2L)2 

1  +  ~~^ 

36,ooor2 

and  if  a  factor  of  safety  of  4  is  used  S  becomes 

12,500 
S=   " 


r  36,ooor2 

in  which  L  is  the  length  in  feet  and  r  is  the  "radius  of  gyration"  in 
inches.  Gordon's  formula  is  the  oldest  and  best  known  of  all  those 
used  in  modern  practice  and  is  found  in  the  handbooks  under  the 
heading  "Safe  loads  .  .  .  for  .  .  .  Columns." 

The  formulas,  giving  S  as  required  by  the  building  laws  of  dif- 
ferent cities,  vary  from  Gordon's  formula.  The  table  of  "Allowable 
Unit  Stress  and  Loads"  is  not  indexed  but  can  be  found  in  the 
1909  edition  of  Cambria — on  page  310.  It  gives  S  as  required 
by  the  New  York  Building  Code  as  15,200  —  58  l/r,  both  /  and  r 
being  measured  in  inches.  This  formula  is  much  more  simple  than 
Gordon's. 

To  design  a  column  to  conform  to  the  requirements  of  the  New 
York  Building  Law,  it  is  necessary  to  understand  what  /  and  r  stand 
for.  /  is  the  length  in  inches  of  the  unsupported  length  of  the 
column,  this  distance  usually  being  taken  as  the  height  from  the 
top  of  the  floor  beam  to  the  bottom  of  the  floor  beam  directly  above 
it.  r  is  the.  least  radius  of  gyration  and  this  term  needs  some 
explaining. 

As,  in  the  second  article,  there  was  no  definition  of  the  moment  of 
inertia  given,  the  architect  need  not  bother  to  find  a  definition  of 
the  least  radius  of  gyration.  The  only  important  thing  to  remember 
is  the  formula  r2  =  I  /a  in  which  r  is  the  radius  of  gyration,  7  is  the 
moment  of  inertia,  and  a  is  the  area  of  the  cross  section.  To  find  r 
it  is  necessary  to  first  find  7. 

Take  a  problem  in  which  it  is  necessary  to  design  a  plate  and 


40        ENGINEERING  FOR  ARCHITECTS 

channel  column,  having  a  section  strong  enough  to  withstand  a  load 
of  300  tons,  and  having  an  unsupported  length  of  fourteen  feet. 
The  first  step  is  to  assume  a  trial  section. 

It  has  been  found  that  for  heights  from  twelve  to  sixteen  feet 
and  average  column  sections,  S  is  approximately  12,000  pounds  or 
six  tons  per  square  inch.  As  the  load  (P)  is  300  tons  and  S  is  6 
tons  per  square  inch  A  must  be  300  -f-  6  =  50  square  inches. 

Assuming  that  the  architect  desires  that  the  column  shall  not 
be  more  than  fourteen  inches  in  any  direction,  this  limits  the  de- 
signer to  the  use  of  twelve-inch  channels  and  fourteen-inch  plates. 
Good  design  usually  results  from  having  as  nearly  equal  areas  in 
the  plates  and  channels  as  possible.  To  get  an  approximate  area 
for  each  channel,  divide  50  square  inches  by  four  and  12.5  square 
inches  is  the  required  area.  A  twelve-inch  channel,  weighing  forty 
pounds  per  foot,  will  have  an  area  of  11.76  square  inches.  This  is 
the  largest  twelve-inch  channel  there  is,  and  will  have  to  be  used 
in  the  design  of  the  columns.  Two  channels  will  have  an  area  of 
23.52  square  inches.  50.00  -  23.52  =  26.48  square  inches  to  be  made 
up  of  plates.  There  will  be  two  sets  of  plates,  each  having  an  area 
of  26.48  -j-  2  =  13.24  square  inches.  As  the  width  of  the  plates  is 
fourteen  inches,  the  thickness  must  be  13.24^-  14,  which  equals 
roughly  one  inch.  The  trial  section  then  will  be  composed  of  two 
twelve-inch,  forty-pound  channels  and  two  fourteen  by  one-inch 
plates. 

In  order  to  find  the  moment  of  interia  of  this  section,  the  loca- 
tion of  the  channels  and  plates  with  relation  to  the  center  lines  must 
be  established.  Under  the  heading  "  Standard  Spacing  of  Rivet  arid 
Bolt  Holes"  in  the  handbooks,  the  distance  " m"  from  the  back  of 
the  channel  to  the  center  line  of  the  rivet  hole  in  the  flange  is  given 
for  nearly  all  sizes  of  channels.  The  distance  for  a  twelve-inch 
forty-pound  channel  in  2j".  In  the  fabrication  of  a  column  the 
distance  from  the  edge  of  the  plate  to  the  center  of  the  rivet  hole 
is  usually  one  inch  and  a  half.  This  makes  the  distance  from  the 
edge  of  the  plate  to  the  back  of  the  channel  2\"  +  ij"  =  3! "  (Fig.  32). 
The  plate,  being  fourteen  inches  wide,  the  distance  from  the  edge 
to  the  center  is  seven  inches  and  the  distance  from  the  back  of  the 
channel  to  the  center  will  be  7"  -  3$"  =  3i" '. 

The  moment  of  inertia  shown  in  Fig.  32  around  axis  XX  will 
be  different  from  that  around  YY,  and  it  will  be  necessary  to  find 
both. 


ENGINEERING  FOR  ARCHITECTS 


In  the  handbooks,  under  the  heading  of  "Properties  of  Standard 
Channels"  the  channels  are  shown  to  have  one  axis  marked  i-i 
and  another  marked  2-2.  /,  for  axis  i-i  of  a  twelve-inch,  forty- 
pound  channel,  is  found  to  be  196.9.  For  two  channels  this  moment 
of  inertia  is  393.8.  To  find  the  moment  of  inertia  of  the  column 
section,  the  /  for  the  plates  around  XX  must  be  found  and  added 


to  that  of  the  channels  to  give  the  total  moment  of  inertia.  In  order 
to  find  this,  the  formula  1=1'+  ah2  must  be  used.  I'  equals  the 
moment  of  inertia  of  the  plates,  a  equals  their  area,  and  h  equals  the 
distance  from  the  axis  XX  to  their  center  of  gravity.  In  Fig.  32 
h  is  found  to  be  6"  +  \"  =  6|".  a  equals  I4"xi"=i4  square 
inches.  The  only  other  factor  to  be  found  is  /'  and  this  is  determined 
by  the  formula  7  =  TV  X  bd*.  b  =  14",  d  =  i",  so  /  X  TV  X  14  X 
I  X  i  X  I  =  1.16.  The  moment  of  inertia  of  the  plates  around  XX 
is  given  as  /=  1.16  +  (14  x  6J  X  6J)  =  1.16  +  591.5  =  592.6.  As 
1.16,  or  /',  is  such  a  small  quantity  in  relation  to  591.5,  it  is  often 
disregarded  by  engineers.  There  are  two  sets  of  plates,  so  the  mo- 


42  ENGINEERING  FOR  ARCHITECTS 

ment  of  inertia  of  the  two  sets  will  be  592.6  X  2=  1185.2.  Add 
this  to  the  /  of  the  two  channels  and  the  total  /  for  the  section  will 
be  /=  393+  1152=  1545. 

To  find  the  moment  of  inertia  around  YY,  the  same  method  is 
employed.  The  /  of  one  set  of  plates  is  given,  as  before,  by  the 
formula  /  =  yV  X  bds,  only,  in  this  case,  b=  i"  and  d  =  14". 
/  =  y1^  x  I  X  14  X  14  X  14  =  229.  Two  sets  of  plates  will  have  a 
moment  of  inertia  of  458.  The  /  for  the  channels  will  be  determined 
by  the  formula  I  =  I'  +  ah2.  In  the  tables  /'  for  a  twelve-inch, 
forty-pound  channel,  around  axis  2-2  is  6.63.  The  area  of  each 
channel  is  11.76.  h  is  the  distance  from  axis  2-2  to  the  center  line 
YY.  The  distance  from  2-2  to  the  back  of  the  channel  is  given  in 
the  handbook  is  .72".  We  have  found  that  the  distance  from  the 
back  of  the  channel  to  the  center  is  3.25  inches,  so  the  distance 
from  YY  to  2-2  is  3.25"  +  .72  =  3.97  inches.  To  find  the  /  of  each 
channel  around  YY.,  simply  substitute  in  the  formula  I  =  6.63  + 
11.76x3.97x3.97=  192.  Two  channels  will  have  a  moment  of 
inertia  of  384  anfl  this,  added  to  the  /  of  the  plates,  will  give 
384+458=  842. 

The  two  moments  of  inertia  are  1545  and  842.  If  one  had 
equaled  the  other,  the  design  would  have  been  better,  but  the  archi- 
tectural requirements  make  this  impossible  and  the  column  will 
have  a  greater  tendency  to  fail  around  axis  YY  than  around  XX. 

The  only  reason  for  finding  /  for  the  section  is  for  the  purpose 
of  determining  the  least  radius  of  gyration.  As  the  area  of  the 
section  also  enters  into  the  calculation  this  must  be  determined. 
The  channels  have  together  an  area  of  23.52  square  inches,  and  the 
two  sets  of  plates  will  have  an  area  of  28  square  inches,  making 
a  total  area  of  51.52  square  inches.  As  r2  =  \,  r2  =  16.4,  and  r 
equals  the  square  root  of  this  or  4.05. 

This  value,  4.05,  is  the  least  radius  of  gyration  and  can  be  used 
to  find  the  value  of  S  in  connection  with  the  above  problem.  S  = 
15,200  -  58  7  or  15,200  -  58  X  14  X  12  -f-  4.05  =  12,800  pounds  per 
square  inch  approximately. 

As  the  area  of  the  section  is  51.5  square  inches,  and  the  allowable 
bearing  value  for  the  section  is  12,800  pounds  per  square  inch,  this 
gives  the  total  strength  of  the  columns  as  12, 800  X  51.5  =  659,200 
pounds  or  about  330  tons.  This  is  too  large,  and  a  reduction  in  the 
area  of  the  column  section  must  be  made.  Instead  of  using  one- 
inch  plates,  seven-eighth-inch  plates  can  be  made  use  of. 


ENGINEERING  FOR   ARCHITECTS  43 

As  a  check  for  this  again  find  the  moment  of  inertia  around  YY. 
The  channels  will  give  the  same  /  as  before  —  384  —  and,  looking 
in  the  handbooks  under  the  heading  of  "Moments  of  Inertia  — 
Tables"  the  /  for  a  plate  14"  X  f"  is  found  to  be  200.08.  Two 
plates  will  have  an  /  of  400.16.  The  total  /  for  the  section  is  384  + 
400  =  784.  The  area  is  23.52  +  24.50  =  48  square  inches  —  roughly 
—  and  r2  =  V-g*  =  16.3.  r  =  4.03.  S=  12,776  and  the  strength 
of  the  section  is  12,776  x  48  =  613,248  pounds  or  307  tons  which  is 
safe. 

If  it  is  desired  to  use  Gordon's  formula,  as  given  in  the  hand- 
books, it  will  be  found  that  the  heavier  section  will  be  required. 

12,500 


1  36,ooor2 
L  =  14  and  r2  =  16.4  so 

12,500 

~i  + 


36,000X16.4 

The  area  being  51.5   square  inches,  the  strength  of  the  section   is 
11,920  x  51.5  =  615,000  pounds  or  307  tons. 

In  the  Cambria  handbook  safe  loads  are  given  in  thousands  of 
pounds  for  various  column  sections.  The  plate  and  channels 
columns  found  in  the  tables  have  plates  varying  in  thickness  from 
J"  to  f "  or  from  f "  to  f " '.  In  no  case  is  there  a  heavy  enough  section 
having  twelve-inch,  forty-pound  channels  and  fourteen-inch  plates 
to  carry  the  3OO-ton  load.  By  proportion,  however,  it  is  possible 
to  check  the  result  given  above. 

From  the  tables  —  page  270  in  the  1909  edition  of  Cambria  - 
it  can  be  found  that  a  column  having  a  section  made  of  two  twelve- 
inch,  forty-pound  channels,  two  J"x  14"  plates,  and  having  a  clear 
height  of  fourteen  feet,  will  support  a  load  of  365,000  pounds.  The 
same  section  having  \"  X  14"  plates  will  support  a  load  of  448,000 
pounds.  The  addition  of  \"  to  the  plates  will  give  an  increase  of 
supporting  power  of  83,000  pounds.  If  this  is  true,  an  increase  of 
\"  would  give  an  increased  supporting  value  of  166,000  pounds. 
If  the  section  having  \"  plates  will  support  448,000  pounds  a  section 
having  i"  plates  will  support  448,000+  166,000=  614,000  pounds 
or  307  tons.  This  checks  with  the  results  given  above,  but  as  this 
method  is  extremely  rough,  it  should  only  be  used  in  checking. 


44  ENGINEERING  FOR  ARCHITECTS 

The  values  given  in  the  handbooks  for  safe  loads  on  columns 
do  very  well  for  light  loads,  but  when  heavy  loads  are  encountered 
and  the  area  of  the  column  is  limited,  it  is  necessary  to  figure  the  sizes 
of  the  members  in  the  cross  section.  However,  if  there  were  no  limit, 
the  sizes  can  be  taken  directly  from  the  handbooks.  It  will  be 
found  that  a  column  composed  of  two  15"  channels,  weighing  45 
pounds  per  foot,  and  two  17"  X  y-J-"  plates  will  support  a  load  of 
606,000  pounds  for  a  clear  height  of  14  feet.  Gordon's  formula, 
as  used  in  all  these  cases,  gives  a  heavier  section  than  is  absolutely 
necessary  and  the  New  York  Building  Code  formula  gives  more 
economical  results. 

In  the  cases  mentioned  in  this  article,  plate  and  channel  columns 
alone  have  been  used.  This  is  done  because  for  a  beginner  the 
section  given  by  the  use  of  plates  and  channels  has  several  advantages. 
It  is  almost  a  square  section  and  therefore  can  be  turned  in  such  a 
direction  as  will  suit  the  architectural  requirements,  without  materi- 
ally affecting  the  design.  The  riveting  of  plate  and  channel  columns 
is  very  simple  but  connections  are  not  made  easily. 

Columns  are  fabricated  in  two  story  lengths.  The  advantages 
of  this  practice  are  obvious  —  the  frame  of  the  building  is  made 
stiffer  and  the  fabrication  and  erection  is  simplified.  The  loading 
on  the  column  increases  at  each  floor,  and  engineers  usually  figure 
the  size  of  each  two-story  length  so  that  it  will  withstand  the  heaviest 
load  coming  upon  it.  If  the  architect  has  a  column  schedule  in  his 
office,  it  would  be  well  for  him  to  consult  this,  noticing  the  loads 
brought  to  each  column,  the  height  of  each  length,  and  the  areas 
of  the  sections. 

In  skeleton  steel  construction,  the  columns  bring  the  loads  of 
the  building  down  to  the  foundations.  Usually  these  loads  are  dis- 
tributed over  the  foundations  by  means  of  grillage  beams  and  the 
determination  of  the  sizes  of  grillage  beams  will  be  taken  up  in  the 
next  chapter. 

Further  consideration  of  columns  will  be  found  in  Chapter  IX. 


CHAPTER  VI 

Grillage  beams.  Failure  by  crushing.  Assumed  number  of  beams  in  lower  layer. 
Calculated  lengths.  Size  of  concrete  footing.  Checks.  Grillage  for  outside 
columns. 

THE  subject  of  grillage  foundations  embraces  theoretical  and 
practical  considerations  too  numerous  to  be  discussed  within 
the  limits  of  a  single  chapter.  The  best  that  can  be  done  is  to  give 
a  general  description  of  this  type  of  foundation  and  to  show  how 
engineers  dispose  of  some  of  the  conditions  involved  in  its  design. 

The  Gothic  builders  first  worked  on  the  principle  of  concentrat- 
ing the  weight  of  their  edifices  upon  isolated  supports.  By  means 
of  balancing  thrust  with  counter-thrust  the  weights  of  vaults,  roofs, 
and  walls  were  transferred  to  large  piers,  and  the  piers  carried  the 
loads  to  the  masonry  footings.  In  modern  buildings  steel  columns 
serve  the  same  purpose  as  the  isolated  piers,  and  the  problem  of 
foundation  design  arises  in  distributing  the  loads,  brought  to  the 
footings,  over  an  area  large  enough  so  that  the  weight  per  square 
foot  will  not  exceed  the  bearing  value  of  soil  or  concrete. 

The  average  steel  column  is  never  much  more  than  a  foot  and 
a  half  square,  and  carries  from  300  to  500  tons.  A  square  foot  of 
reinforced  concrete  has  a  unit-bearing  value  of  35  tons,  and  soil  can 
only  support  4  tons  per  square  foot.  It  can  be  seen  from  this  that  a 
load  brought  to  the  foundation  by  two  and  a  quarter  square  feet  of 
column,  must  be  distributed  over  many  more  square  feet  of  con- 
crete or  soil,  or  the  footing  will  not  bear  up  under  the  load.  Grillage 
beams  are  used  for  this  purpose  of  distribution. 

Fig.  33  shows  a  perspective  view  of  a  grillage  foundation.  Under 
the  column  is  a  steel  slab  17  inches  wide,  3  inches  thick  and  i  foot 
10  inches  long.  Under  this  are  placed  six  1 5-inch  channels,  ar- 
ranged back  to  back  in  three  pairs,  and  under  the  channels  are  placed 
seven  I-beams  which  distribute  the  5oo-ton  load  over  the  reinforced 
concrete. 

In  all  previous  articles,  the  only  method  used  in  determining 
the  sizes  of  beams  to  withstand  certain  loads  was  to  supply  a  beam 
strong  enough  to  resist  the  tendency  to  fail  by  bending.  Although 
it  is  always  necessary  to  check  the  sizes  of  grillage  beams,  to  deter- 


ENGINEERING  FOR  ARCHITECTS 


mine  their  ability  to  resist  bending,  other  considerations,  such  as 
crushing  of  the  web,  govern  the  design. 

As  in  the  case  of  column  design,  it  is  necessary  to  make  trials  in 
order  to  determine  the  actual  sizes  of  the  grillage  beams,  and  the 
general  dimensions  of  the  footings.  The  dimensions  of  the  slab  are 
governed  by  those  of  the  column.  A  column,  made  of  1 2-inch  chan- 
nels and  1 6-inch  plates,  with  4-inch  by  6-inch  clip  angles,  requires  a 


FIGURE  33 

plate  17  inches  wide  by  i  foot  10  inches  long.     The  method  of  deter- 
mining the  thickness  of  the  plate  will  be  taken  up  later. 

When  the  foundations  rest  upon  bed  rock,  the  outside  dimensions 
of  the  footing  are  governed  by  the  bearing  power  of  the  reinforced 
concrete.  The  New  York  Building  Department  allows  a  pressure 
of  one  quarter  of  a  ton  (500  Ibs.)  per  square  inch  on  that  reinforced 
concrete  which  is  in  direct  contact  with  the  steel.  In  other  words, 
the  portion  of  concrete  included  between  the  flanges  of  the  lower 
layer  of  beams  (x  —  Fig.  34)  cannot  be  considered  as  supporting  any 
portion  of  the  500  tons.  For  this  reason  it  is  necessary  to  use  a 


ENGINEERING  FOR   ARCHITECTS 


large  number  of  light  beams,  having  comparatively  large  flanges, 
instead  of  a  few  heavy  beams  with  comparatively  small  flanges. 

As  a  trial  seven  twelve-inch,  thirty-five-pound,  I-beams  will 
be  used  in  the  lower  layer.  These  beams  have  flanges  5.09 
inches  wide  and  a  portion  of  one  beam,  one  foot  long,  will  have 
an  area  of  12"  X  5.09"  =  61  square  inches.  Seven  beams  will  have 
61  X  7  =  427  square  inches  of  flange  area  bearing  on  the  concrete 
per  lineal  foot  of  beam.  As  each  square  inch  of  concrete  will  sup- 


FIGURE  34 

port  one  quarter  of  a  ton,  427  square  inches  will  support  427  -=-  4  = 
107  tons.  There  are  500  tons  to  be  carried  so  500  -=-  107  =  4.6  feet 
of  beams  will  be  necessary.  To  give  an  even  figure  these  beams  will 
be  made  four  feet  six  inches  long. 

It  is  usually  considered  good  practice  to  have  the  footing  square 
so  the  upper  layer  of  beams  will  be  made  four  feet  six  inches  long  also. 
The  concrete  should  project  at  least  three  inches  beyond  the  ends 
of  the  beams,  so  the  outside  dimensions  of  the  footing  will  be  five 
feet  by  five  feet. 

We  now  have  assumed  the  area  of  the  slab  and  the  area  of  the 
concrete  footing  and  have  assumed  the  sizes  of  beams  in  the  lower 


48 


ENGINEERING  FOR   ARCHITECTS 


layer.  For  the  given  conditions  it  is  necessary  to  determine  the 
sizes  of  beams  in  the  upper  layer.  These  beams  will  have  to  with- 
stand a  load  of  500  tons  coming  down  on  17  inches  of  their  length. 
On  account  of  the  dimensions  of  the  slab  there  can  be  only  three 
beams,  so  each  beam  will  have  to  support  500  -f-  3  =  166.6  tons. 
This  load  will  tend  to  crush  the  web  of  the  beam,  and  we  must  supply 
a  beam  strong  enough  to  resist  this  crushing.  The  number  of  square 
inches  of  steel  in  the  web  of  each  beam  under  the  slab  will  be  given 
by  17"  X  t,  in  which  "t"  is  the  thickness  of  the  web.  If  each  square 
inch  will  support  8  tons — the  safe  crushing  value  of  steel —  the  total 

bearing  value  in  tons  of  each  web  is  17 
X  t  X  8.  This,  of  course,  must  equal  166.6, 
or,  17x^x8  =  166.6.  t  must  equal  166.6 
-=-  136=  1.2  inches.  There  is  no  I-beam 
having  a  web  as  thick  as  this,  so  in  place 
of  a  single  I-beam,  two  1 5-inch,  45-pound 
channels  riveted  back  to  back  are  used,  the 
two  webs  having  a  combined  thickness  of 
1.24  inches.  The  upper  tier  of  beams  will 
then  consist  of  six  1 5-inch,  45-pound  chan- 
nels, 4  feet  6  inches  long.  The  only  other 
dimension  to  be  found  in  this  trial  footing 
is  the  thickness  of  the  plate. 

Fig.  35  shows  a  plan  of  the  column  and 
plate.  The  axis  XX  divides  the  column 
into  two  equal  parts  and  a  load  of  250  tons  is 


FIGURE  35 


brought  down  on  the  right,  and  an  equal  one  on  the  left,  of  this  axis. 
The  neutral  axis  of  the  right  section  of  the  column  is  assumed  to  be 
6  inches  from  XX  and  the  load  of  250  tons  will  be  concentrated  along 
this  line.  The  center  lines  of  the  beams  under  the  slab  are  shown 
by  dotted  lines  7!  inches  apart.  The  upward  load  upon  the  slab 
exerted  by  each  beam  will  be  500-7-  3  =  166.6  tons.  The  upward 
moment  will  be  166.6x71=1291  inch-tons.  The  downward 
moment  will  be  250  X  6  =  1,500  inch-tons.  The  difference  is  1500  — 
1291  =  209  inch-tons  as  the  bending  moment  set  up  in  the  slab. 

Referring  to  the  first  chapter  we  find  that  M  =  S  I/c.  S  =  8  tons, 
and  I/c  =1/6  bd?.  In  this  case  b  =  17  inches,  and  the  problem 
arises  in  determining  d.  209  =  8  x  i  X  17  X  d2.  d?  =  9,  so  d  =  3 
inches.  This  gives  us  all  the  dimensions  of  the  trial  footing  —  the 
size  of  the  plate,  the  sizes  of  the  beams  in  the  upper  and  lower  layer, 


ENGINEERING  FOR  ARCHITECTS 


49 


and  the  dimensions  of  the  concrete  pier.  It  is  necessary  to  check 
the  beams,  however,  to  determine  their  ability  to  resist  bending  and 
shearing  stresses. 

In  cases  where  bending  has  been  considered,  the  formula  M  =  \Wl 
has  always  been  used.  A  modification  of  the  formula  is  used  in  the 
case  of  grillage  beams.  The  formula  takes  the  form  M  =  \W  (I  -  a) 
in  which  (/  —  a)  is  the  overhanging  length  of  the  beams.  In  Fig. 
34,  "a"  is  the  length  of  the  slab  or  17  inches,  and  "/"  is  the  total 


4-6 


FlGURE    36 

length  of  the  channels  or  4  feet  6  inches.     (/  -  a)  equals  4  feet  6  inches 
minus  I  foot  5  inches  or  3  feet  and  i  inch. 

If  M  =  |  W  (I  —  a)  then  in  the  case  of  the  upper  beams  M  =  f  X 
500x3.1=  194  foot-tons.  The  section  modulus  is  found  when 
foot-tons  are  used  by  multiplying  the  maximum  bending  moment 
by  ij,  as  explained  in  chapter  III.  194  X  J=  291  =  I/c.  This  is 
the  combined  I/c  for  the  six  channels  and  each  channel  will  have 
to  have  a  section  modulus  equal  to  or  greater  than  291  -f-  6  =49. 
This  is  less  than  the  I/c  for  a  1 5-inch,  4O-pound  channel,  which  is 
given  as  50,  so  the  channels  are  -safe  as  far  as  bending  is  concerned. 


50        ENGINEERING  FOR  ARCHITECTS 

The  only  other  way  in  which  failure  can  occur  is  by  shearing  the 
ends  of  the  channels  off  at  the  edges  of  the  plate.  (Fig.  34.)  The 
projection  of  the  channels  beyond  each  side  of  the  plate  is  approxi- 
mately i '-6".  This  is  actually  one  third  of  the  total  length  of  the 
channels.  If  the  channels  must  support  500  tons,  each  projecting 
length  must  support  500^-  3  =  166.6  tons.  Each  channel  will  then 
have  a  maximum  shear  of  166.6  H-  6  =  27.7  tons.  The  area  of  the 
cross  section  of  a  1 5-inch,  45-pound  channel  is  13.24  square  inches, 
and  the  shear  per  square  inch  on  the  section  is  27.77-7-  13.24=  2 
tons,  approximately.  This  is  well  within  the  safe  shearing  value  of 
steel  of  4.5  tons  per  square  inch. 

We  now  know  that  the  channels  in  the  upper  layer  are  safe  and 
we  must  check  the  sizes  of  the  I-beams  in  the  lower  tier.  Fig.  36 
shows  the  location  of  the  channels  and  the  slab  in  relation  to  the 
lower  beams,  "a"  in  this  case  is  i  foot  10  inches,  and  (/ -  a)  must 
equal  4  feet  6  inches  minus  i  foot  10  inches  or  2  feet  8  inches. 
I/c  =  I  X  i  X  500  x  2.66  =  250  for  seven  beams.  250 -=-7  =35. 
The  I/c  for  a  twelve  inch,  thirty-five  pound  I-beam  is  38  so  these 
beams  will  not  fail  by  bending. 

These  beams  might  fail  by  crushing  and  shearing,  however,  and 
it  is  necessary  to  determine  whether  this  is  the  case  or  not.  The 
projecting  ends  of  the  beams  beyond  the  channels  are  i  foot  4  inches 
long  and  we  will  be  safe  in  assuming  that  the  total  shearing  stress 
is  150  tons.  As  there  are  seven  beams,  the  shear  per  beam  will  be 
150 -j-  7  =  2I-4  tons.  The  area  of  the  cross  section  of  twelve  inch, 
thirty-five  pound  I-beam  is  10.29  square  inches.  So  the  shear  per 
square  inch  will  be  21.4  -*•  10.29  =  2  tons  approximately  which  is  safe. 

The  section  of  the  web  of  one  beam  directly  under  the  slab  has 
an  area  of  22  inches  long  by  .44  inches  thick  or  9.8  square  inches. 
Seven  beams  will  have  a  total  area  of  9.8  X  7  =  68.6  square  inches. 
As  each  square  inch  will  withstand  a  pressure  of  8  tons,  68.6  square 
inches  will  support  68.6  X  8  =  548.8  tons,  so  the  beams  will  safely 
support  the  5OO-ton  load. 

The  trial  footing,  therefore,  is  safe  in  every  particular. 

The  conditions  accounted  for  so  far,  have  been  for  footings  where 
it  was  possible  to  have  the  concrete  rest  on  bed  rock,  and  where  there 
was  no  limit  to  the  dimensions  of  the  footing  in  any  direction.  This 
is  only  one  of  many  conditions  that  may  arise.  In  case  the  column 
is  an  outside  one  and  placed  about  two  feet  back  of  the  building  line 
it  would  be  impossible  to  have  a  base  five  feet  square.  This  condi- 


ENGINEERING  FOR  ARCHITECTS 


tion  is  shown  in  Fig.  37,  and  the  footing  should  be  longer  than  it  is 
broad. 

In  this  case  the  column  having  a  heavy  wall  load  to  support  will 
have  a  load  of  1,061  tons.  The  size  of  the  footing  will  be  considered 
first.  As  the  column  center  line  is  two  feet  back  of  the  building  line 
the  total  width  of  the  footing  cannot  be  more  than  four  feet  or  forty- 
eight  inches.  If  five  beams  are  used  in  the  lower  tier  then  there  will 
be  four  spaces  between  them,  and  a  slight  over-hang  on  each  end,  so 
the  beams  will  be  9''  on  centers.  The  average  width  for  heavy 
flanges  is  about  seven  inches  so  a  foot  of  beam  will  have  12  X  7  =  84 
square  inches  of  flange  bearing  on  the  concrete,  5  beams  will  have 

T3v i  LP i  rtc  l_i ncj  *,lah    p^t'x^Lf  r2-lt  " 


.* 

\  \ 

*") 

y 

—  — 

li 

i 

1  \- 

sii'-ttK;*''**?}**' 

•                           to  o~                      T 

FIGURE  37 

420  square  inches.  Each  square  inch  of  concrete  will  stand  one- 
quarter  of  a  ton  so  420  -=-  4  =  105  tons  per  foot  of  beam.  1,061  -=-  105 
=  10  feet.  In  the  trial  footing  the  beams  in  the  upper  tier  will  be 
taken  as  4  feet  long,  and  those  in  the  lower  tier  will  be  10  feet  long. 
The  conditions  of  fabrication  will  make  the  slab  24  inches  wide  by 
2  feet  1 1  inches  long. 

The  next  step  is  to  determine  the  sizes  of  the  beams  that  make 
up  the  footing.  The  four  beams  in  the  upper  layer  are  more  apt 
to  fail  by  crushing  than  by  bending  on  account  of  their  short  length. 
The  slab  covers  2  feet  of  the  beam  so  the  thickness  of  the  web  must 
be  found  as  follows:  1,061  =  24  x  /  X  4  X  8  or  t  =  1061/768  =1.4 
inches,  so,  for  resistance  to  crushing,  eight  1 5-inch  channels  weigh- 
ing 50  pounds  per  foot  must  be  used.  The  shear  is  one-quarter  of 
the  total  load  as  the  projecting  length  of  the  channels  on  either  side 
of  the  slab  is  one-quarter  of  the  total  length.  1,061  +  4  =  265  tons. 
In  this  case,  because  some  engineers  figure  that  only  the  web  should 
resist  the  shear,  we  will  determine  what  the  shear  per  square  inch 
on  the  web  of  the  channels  will  be.  The  total  area  of  the  web  is 


52        ENGINEERING  FOR  ARCHITECTS 

8  X  15  X  .72  =  86  square  inches  =  265  -r-  86  =  3  tons  per  square  inch. 
The  beams  are  safe  from  failure  by  shearing. 

Failure  by  bending  will  be  considered  next.  I/c=  i^M  =  ij 
X  i  X  W  (/  -  a)  =  3/2  X  |  X  1 ,06 1  X  2  =  397.  This  is  the  combined 
section  modulus  for  eight  channels.  I/c  for  one  channel  is  397  +  8  = 
50.  The  section  modulus  of  a  1 5-inch,  5o-pound  channel  is  53.7  so 
the  upper  layer  of  beams  is  safe  in  every  respect. 

It  would  seem  that  the  long  beams  in  the  lower  tier  would  have 
a  tendency  to  fail  by  bending.  (/  -  a)  in  this  case  equals  10  feet  minus 
3  feet,  or  7  feet.  The  total  I/c  will  equal  3/2  X  f  X  1,061  X  7  =  1,393. 
The  section  modulus  for  each  beam  is  1,393  -z-  5  =  278. 

There  is  no  standard  beam  that  is  heavy  enough  to  withstand  this 
bending,  so  a  heavy  beam  reinforced  with  plates  on  the  top  and 
bottom  flanges  will  be  necessary.  In  case  24-inch,  loo-pound  I-beams 
are  used,  these  plates  will  be  one  inch  thick.  The  method  of  deter- 
mining this  thickness  will  be  taken  up  in  a  later  chapter.  To  resist 
crushing  the  web  must  have  a  thickness  of  1,061  -=-36x5x8=  .73 
and  the  web  of  a  24-inch,  loo-pound  beam  will  be  large  enough. 
These  beams  should  also  be  checked  for  shear  but,  as  we  have  en- 
countered no  tendency  toward  failure  in  this  direction,  it  can  be 
assumed  that  there  is  none  in  this  case. 

When  columns  rest  on  soil  foundations  other  problems  arise,  and, 
when  the  dimensions  of  the  footing  are  limited  in  two  direction0,  as 
in  the  case  of  a  corner  column,  still  further  considerations  have  to  be 
taken  care  of.  These  complications  will  be  dealt  with  in  the  next 
chapter. 


CHAPTER  VII 

Grillage  foundations  on  soil.  Size  of  concrete  footing.  Length  of  beams.  Maxi- 
mum bending  moment  and  section  .modulus  of  upper  beams.  Tests  for  shearing 
and  crushing.  Design  of  lower  layer  of  beams.  Footings  for  corner  columns. 
Cantilever  beams. 

WHEN  column  loads  have  to  be  distributed  over  soil  the  methods 
employed  in  figuring  the  footings  are  the  same  as  in  Chapter 
VI,  but,  as  soil  will  only  withstand  a  compressive  force  of  four  tons  per 
square  foot,  the  results  differ  from  those  of  the  preceding  chapter. 
In  the  case  where  there  was  a  bearing  on  rock,  the  failure  of  the 
grillage  might  have  occurred  in  the  concrete  under  the  steel,  the  con- 
crete being  weaker  than  the  rock.  For  this  reason  it  was  necessary 
to  reinforce  the  concrete  and  to  assume  that  only  the  concrete  in 
direct  contact  with  the  steel  had  any  bearing  value.  These  precau- 
tions will  not  be  necessary  when  the  bearing  is  on  soil  as  the  earth 
will  settle  before  the  concrete  will  fail.  In  Chapter  VI,  the  first 
consideration  was  the  size  of  the  concrete  slab.  In  the  present  case 
it  is  the  area  of  the  soil  over  which  the  concrete  slab  is  spread  that 
must  be  determined  first. 

Given  a  column  load  of  400  tons,  and  the  safe  bearing  value  of 
soil  as  4  tons  per  square  foot,  the  area  over  which  the  load  is  to  be 
distributed  is  400  -^-  4  =  100  square  feet.  If  the  footing  is  made 
square  the  outside  dimensions  will  be  10  feet  by  10  feet.  (Fig.  38.) 

Allowing  a  projection  of  concrete  of  4  inches  on  all  sides  the 
length  of  the  beams  in  both  layers  will  be  10  feet,  minus  8  inches,  or 
9  feet  4  inches.  (9.33  feet.) 

The  steel  slab  under  the  column  will  be  made  3  feet  square,  and, 
using  the  formula  for  bending  given  in  Chapter  VI,  the  maximum 
bending  moment  for  the  beams  in  the  upper  layer  will  be 

M  =  |  X  400  X  (/  -  0),  or,  50  X  6.33  =  316  foot-tons. 

The  section  modulus  is  316  X  ij  =  474.  It  is  possible  to  place  three 
beams  under  the  slab  and  the  sections  modulus  for  each  beam  will 
be  474 -^  3  =  158.  Three  24-inch,  8o-pound  I-beams  will  be  suffi- 
ciently strong  to  withstand  the  tendency  to  fail  by  bending. 


54 


ENGINEERING  FOR   ARCHITECTS 


The  beams  project  on  each  side  beyond  the  slab  approximately 
one  third  of  their  total  length  and  the  maximum  shear  will  be 

|  X  400  =133  tons. 

The  web  of  a  24-inch,  8o-pound  I-beam  is  J  inch  thick,  and  the  area 
of  the  web  is  24  X  \  =  12  square  inches.  Three  beams  will  have  a 
total  web  area  of  12  X  3  =  36  square  inches.  As  the  total  shear  is 
133  tons,  the  shear  per  square  inch  of  web  will  be  133  -r-  36  =  3.66  tons. 
The  Building  Department  allows  a  shear  of  4.5  tons  on  steel  and  the 
beams  are  safe  as  far  as  failure  by  shearing  is  concerned. 

The  only  other  method  of  failure  to  be  considered,  for  the  beams 
on  the  top  tier,  is  the  crushing  of  the  web.     The  slab  extends  over  a 


FIGURE  38 

length  of  three  feet  and  the  webs  are  one-half  an  inch  thick  so  the 
area  of  the  web  under  the  plate  is  36  x  \  X  3  =  54  square  inches. 
The  load  is  400  tons,  and  the  compression  per  square  inch  is  400  -r- 
54  =  7.4  tons,  which  is  safe.  The  upper  layer  of  beams  is  strong 
enough  in  every  particular. 

In  the  lower  layer  the  beams  should  be  placed  so  that  there  will 
be  comparatively  little  space  between  the  adjacent  flanges.  If 
twelve  beams  are  used,  spaced  9!  inches  on  centers,  they  will  prac- 
tically cover  the  concrete.  The  total  section  modulus  for  all  of  these 
beams  will  be  the  same  as  for  those  in  the  upper  layer,  as  W  and 
(/  -  a]  remains  the  same.  This  should  be  checked.  In  order  to  find 
I/c  for  a  single  beam,  simply  divide  474  by  12  and  the  result — 39.5 
— will  be  the  section  modulus.  The  lower  tier  of  beams  will  consist 
of  12  12-inch,  40-pound  I-beams,  9  feet  4  inches  long  and  spaced 
9!  inches  on  centers.  By  the  methods  already  employed  these 


ENGINEERING  FOR  ARCHITECTS 


55 


beams   should   be  checked   for  shearing  and   crushing  of  the  web. 
They  will  be  found  to  be  safe. 

Fig.  39  shows  a  condition  in  which  the  corner  column  is  so  placed 
that  a  square  footing  cannot  be  used  as  two  sides  will  be  outside 
the  building  lines.  As  this  is  not  allowed  in  most  cities,  a  type  of 
foundation  must  be  employed  somewhat 
different  from  any  we  have  considered  so 
far. 

If  the  footing  were  spread,  as  shown  in 
Fig.  39a,  so  as  to  throw  the  center  of  the 
column  off  the  center  of  the  concrete  base 
the  tendency  would  be  to  make  the  footing 
settle  unevenly.  This  tendency  toward 
uneven  settlement  can  be  illustrated  by 
considering  a  row  boat  in  which  the  only 
occupant  is  located  in  the  bow.  The  bow 
will  sink  slightly  in  the  water  and  the  stern 
will  rise.  If  another  person  should  be  | 
placed  in  the  stern  the  two  occupants  { 
would  counter-balance  each  other  and  the 
boat  would  set  evenly.  In  like  manner,  if 
the  footing  of  the  corner  column  were 
extended  far  enough  to  the  left  to  allow 


Bulletin 


I 


FIGURE  39 


i 


FIGURE  393 

the  next  outside  column  to  rest  upon  it,  the  second  column  would 
act  as  a  counter-balance  for  the  corner  one,  and,  provided  the  dimen- 
sions of  the  footing  were  properly  proportioned,  there  would  be  no 
more  tendency  to  settle  in  one  direction  than  in  another. 

Fig.  ^o  shows  a  diagrammatic  plan  and  elevation  of  this  condi- 
tion. Column  No.  I  is  the  corner  column,  having  a  load  of  650  tons, 
and  which  is  located  2  feet  back  of  the  building  lines.  Column  No.  2 
carries  a  load  of  600  tons  and  is  in  line  with  the  corner  column.  The 
distance  between  No.  I  and  No.  2  is  20  feet.  Under  the  two  columns 
two  girders,  Gi  and  G2,  are  placed  and  I-beams  are  distributed  under 
the  girders  to  carry  the  loads  to  the  concrete  footings.  These  foot- 
ings rest  upon  rock. 

If  the  architect  understands  clearly  the  method  of  solving  the 
conditions  referred  to  in  Chapter  VI,  the  only  new  problem  to  be 
solved  is  the  position  of  the  I-beams  and  concrete  footings.  First, 
it  is  necessary  to  assume  the  size  of  the  I-beams  and  determine  their 
length  and  numbers.  As  12-inch,  3i|-pound  I-beams  have  proved 


ENGINEERING  FOR  ARCHITECTS 


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ENGINEERING  FOR   ARCHITECTS  57 

satisfactory  for  most  conditions  imposed  upon  the  lower  layer  of 
grillage  beams,  it  will  be  assumed  that  they  will  be  satisfactory  in 
this  case.  The  length  is  governed  by  the  dimensions  of  the  con- 
crete. As  the  center  of  the  footing  is  2  feet  back  of  the  building  line, 
the  width  of  the  footing  can  only  be  twice  this  or  4  feet.  Allowing 
an  overhang  of  3  inches  of  concrete  beyond  the  ends  of  the  beams, 
the  beams  themselves  can  only  be  3  feet  6  inches  long.  The  flanges 
of  12  inch,  3iJ-pound  I-beams  are  5  inches  wide,  and,  as  the  beams 
are  42  inches  long,  there  will  be  42  X  5  =  210  square  inches  of  flange 
bearing  on  the  concrete  for  each  beam.  In  Chapter  VI  the  rein- 
forced concrete  in  direct  contact  with  the  steel  was  considered  as 
having  a  bearing  power  of  one-quarter  of  a  ton  (500  pounds)  per 
square  inch.  The  concrete  under  each  beam  will  support 

210  -~  4  =  52.25  tons. 

Column  No.  i  has  a  load  of  650  tons.  The  number  of  beams 
under  this  column  will  be  determined  by  dividing  the  load  by  52.25. 
650 -=- 52.25  =  13  approximately.  There  will  be  13  beams  under 
the  corner  column.  In  like  manner  there  will  be  12  beams  under 
column  No.  2.  If  the  beams  are  placed  7J  inches  on  centers  the  set 
under  No.  I  will  spread  over  an  area  of  concrete  of  (12  X  7i)  +  5  =  95 
inches.  The  5  inches  added  to  90  is  the  width  of  a  single  flange. 
Ninety-five  inches  equal  7  feet  n  inches.  (Fig.  41.)  The  I-beams 
under  No.  2  will  cover  7  feet  3!  inches. 

So  far  the  considerations  involved  have  been  exactly  the  same  as 
those  mentioned  in  Chapter  VI.  The  next  step,  however,  is  a  new 
one  and  involves  the  consideration  of  moments.  The  architect 
should  thoroughly  understand  what  is  meant  by  the  term  moment 
and  in  case  there  is  any  doubt  on  the  subject  he  should  look  over 
Chapter  I  again.  A  center  of  moments  can  be  taken  at  any  point, 
and,  for  conditions  of  equilibrium,  the  sum  of  all  moments  around 
this  center  must  equal  zero.  In  other  words,  the  upward  moments 
must  equal  the  downward  moments  around  this  center. 

For  the  given  conditions,  the  center  of  moments  will  be  taken 
at  the  building  line.  Column  No.  I  has  a  load  of  650  tons,  and  the 
distance  from  its  center  to  the  building  line  is  2  feet,  so  the  moment 
around  the  assumed  center  is  650  X  2  =  1,300  foot-tons.  The  second 
column  has  a  moment  of  600  x  22=  13,200  foot-tons.  The  total 
downward  moment  is  1,300+  13,200=  14,500  foot-tons.  To  coun- 
teract this  there  must  be  an  upward  moment  equal  to  it  and  the  two 


58        ENGINEERING  FOR  ARCHITECTS 

sets  of  I-beams  may  be  considered  as  exerting  upward  forces  which 
will  cause  the  upward  moments.  Each  force  will  be  equal  to  that 
brought  down  by  the  column  under  which  each  set  of  I-beams  is 
placed,  and  the  forces  will  be  considered  as  acting  at  their  respective 
centers  of  gravity. 

The  beams  under  column  No.  i  will  be  placed  as  near  the  build- 
ing line  as  possible,  but,  as  there  should  be  a  projection  of  3  inches  of 
concrete  beyond  the  steel,  the  center  of  the  set  will  be  (7'-n"  n-2)  + 
3  =  4/-2J"  from  the  building  line.  The  upward  moment  caused  by  the 
footing  will  be  650  X  4.20=  2735  foot-tons.  Subtracting  this  from 
14,500  foot-tons  leaves  11,764  foot-tons  to  be  taken  up  by  the  second 
set  of  beams.  As  this  moment  must  equal  a  force  multiplied  by  a 


Points  of  no  shear 


3 hear  £>/agram 


FIGURE  41 

distance,  and  the  force  (600  tons)  is  known,  the  question  is  to  deter- 
mine the  distance.  Let  X  equal  this  distance.  Then 

600  x  X  =  11,764,  or  X  =  11,764  -r-  600  =  19.60. 

Fig.  42  shows  a  diagram  representing  these  conditions  —  the  two 
columns  acting  downward  and  the  two  footings  acting  upward  with 
their  respective  lever  arms. 

We  now  have  the  size,  number  and  location  of  the  I-beams  in  the 
trial  footing.  From  knowledge  which  has  already  been  acquired  it 
is  possible  to  lay  out  the  beams  as  shown  in  Fig.  40. 

The  next  step  is  the  determination  of  the  members  in  the  girders. 
The  two  slabs,  under  the  columns,  are  each  2  feet  6  inches  long.  The 
sets  of  I-beams  will  be  assumed  to  act  as  uniform  loads  extending 
over  7  feet  n  inches  and  7  feet  5  inches  respectively.  The  shear 
diagram  is  shown  in  Fig.  41,  and,  as  all  the  loads  are  uniformly  dis- 
tributed, the  shear  is  represented  as  varying  as  a  series  of  sloping 
lines.  In  Chapter  III  the  statement  was  made  that  a  beam  will 
ail  by  bending  at  the  point  where  the  shear  is  zero.  From  the  dia- 


ENGINEERING  FOR  ARCHITECTS        59 

grams  it  can  be  seen  that  the  shear  is  zero  at  any  point  between  A  and 
B.  To  find  the  bending  moment  at  A  we  have  two  forces  to  account 
for  —  the  6oo-ton  load  of  the  column  acting  downward  and  the  600- 
ton  load  of  the  footing  acting  upward.  Both  loads  act  through  their 
centers  of  gravity.  The  moment  of  the  column  load  is  600  tons  X  6.04 
feet  =  3624  foot-tons,  and  the  moment  of  the  footing  is 

600  X  3.65  =  2,190  foot-tons. 

The  difference  is  1434  foot-tons.  In  like  manner,  the  moment 
around  B  is  650  X  (6'-2"  -  3'-ui")  =  650  X  2.208  =  1,435  foot-tons. 
It  will  be  noticed  that  both  moments  are  approximately  equal  and 


Downward   Moments 

600  *2Z-  13,  ZOO   ft-  tons. 
650*    Z  •      1,300    -       •• 

14.,  500     -      "  Total. 

-  Q 


Total, 

FIGURE  42 

also  that  the  same  results  may  be  obtained  by  multiplying  the 
column  load  by  the  distance  from  the  center  of  the  footing;  thus, 
600  tons  X  2.39  =  1434  foot-tons. 

From  the  shear  diagrams  the  maximum  shear  can  be  obtained. 
This  is  found  at  the  point  C  —  the  edge  of  the  slab  under  column  No. 
i  —  and  equals  approximately  400  tons. 

Two  riveted  girders,  made  of  plates  and  angles,  can  be  designed 
according  to  the  principles  mentioned  in  Chapter  IV.  The  shear 
for  two  girders  is  400  tons  and  for  one  is  200  tons.  To  use  the  for- 
mula for  the  depth  of  the  girder  given  in  the  former  article  will  give 
an  excessive  depth  as  the  shear  is  large  in  proportion  to  the  maximum 
bending  moment.  The  most  satisfactory  method  will  be  to  assume 
the  depth  as  some  convenient  figure  —  4  feet  —  and  design  the  girder 
for  this  depth. 

Another  method  is  to  select  the  proper  riveted  girder  from  those 
given  in  the  Carnegie  "Pocket  Companion."  On  pages  from  235  to 
250  are  given  lists  of  plate  girders  the  sizes  of  which  are  determined 


60        ENGINEERING  FOR  ARCHITECTS 

by  the  section  modulus  of  each  girder.  In  the  present  case  the 
maximum  bending  moment  for  two  girders  is  1,435  foot-tons  and  the 
section  modulus  is  1,435  X  ij  =  2,152.  For  one  girder  the  section 
modulus  is  one-half  this  or  1,076.  On  page  249  of  the  Pocket  Com- 
panion, a  girder  will  be  found  having  an  I/c  of  1,083  which  is  the 
nearest  to  the  figure  given  above.  These  girders  will  be  made  of  a 
48-inch  by  J-inch  web  plate,  four  6  x  6  X  J-inch  flange  angles,  and 
two  14  X  f-inch  flange  plates.  In  using  such  a  girder  it  must  be 
remembered  that  the  safe  working  stress  used  as  a  basis  for  its  design 
is  taken  as  8  tons.  Some  building  departments  require  that  for 
riveted  steel  this  stress  should  be  7.0  tons.  The  girder  should  also 
be  stiffened  by  stiffener  angles  directly  under  the  columns  and  at 


t,-  6  5  O.ton  z 

FIGURE  43 

the  points  of  maximum  shear.  The  method  of  designing  such  stif- 
feners  would  be  the  same  as  used  in  Chapter  IV. 

So  far  the  I-beams  in  the  lower  layer  have  not  been  checked  for 
bending,  but  if  (/ —  a)  is  I  foot  and  the  load  on  twelve  I-beams  is 
650  tons  it  will  be  found  that  they  are  safe.  This  is  also  true  regard- 
ing failure  by  shearing  and  crushing. 

Referring  again  to  the  Pocket  Companion,  on  pages  224-25-26-27 
and  28,  it  will  be  found  that  the  problem  of  grillage  beam  design  is 
dealt  with.  In  these  cases  the  footings  are  on  soil  and  for  a  short 
and  comprehensive  treatment  of  the  subject  it  is  probably  the  best 
published. 

For  reasons,  unknown  to  an  engineer,  architects  often  refer  to 
such  girders  as  designed  in  this  article  as  cantilever  girders.  As  a 
matter  of  fact  only  a  small  portion  acts  as  a  cantilever  so  the  bend- 
ing moments  and  shear  are  determined  in  exactly  the  same  manner 
as  if  the  girders  were  simple  beams.  Imagine  Fig.  40  turned  upside 
down  as  shown  in  Fig.  43.  Now  the  column  loads  act  upward  and 
the  uniform  loads  of  the  footings  act  downward.  Only  the  portion 
ab  overhangs  the  support  and  acts  as  a  cantilever.  Because  ab  is  so 
small,  the  bending  set  up  in  the  girders,  due  to  this  projection,  is  not 


ENGINEERING  FOR   ARCHITECTS  61 

enough  to  influence  the  actual  design  of  the  girders  as  the  tendency 
toward  bending  at  the  points  A  and  B  is  so  much  greater. 

As  an  actual  example  of  a  simple  cantilever  beam  note  the  con- 
dition shown  in  Fig.  44.  Here  we  have  a  load  of  15  tons  located  8 
feet  from  R2  and  the  portion  of  the 
beam  projecting  to  the  right  of  R^  acts 
as  a  cantilever. 

The  design  of  such  a  beam  is  simple 
provided  the  architect  does  not  set  diffi- 
culties in  his  own  path.  To  find  the  br^T?2"0" T~772*T~~"I 

reactions  the  same  method  is   employed 

.       ,  r         •       i     L  *r  i  •  FIGURE  44 

as  in  the  case  or  a  simple  beam.      I  akmg 

RI   as   the   center   determine   the   downward   moments   around   this 
center. 

30  tons  X    3  feet  =    90  foot-tons. 
40    "    x    7    "    =  280 
£5    "    X2o    "    =300 
Totals  85  tons  670  foot-tons. 

670  foot-tons  -j-  12  feet  =  55.83  tons  =  R%. 
85  tons  -  55.83  tons  =  29.17  tons  =  RI. 

The  shear  diagram  is  shown  in  Fig.  44  and  from  this  it  is  apparent 
that  there  are  two  points  of  no  shear.  In  other  words  there  are  two 
points  at  which  the  beam  might  fail  by  bending.  It  will  be  necessary 
to  find  the  bending  moment  at  each  point  to  determine  which  is  the 
greater.  The  one  at  a  is  29.17  X  3  =  87.51  foot-tons,  and  the  one  at 
R2  is  15x8=  1 20  foot-tons.  It  is  obvious  that,  if  the  beam  should 
.  fail  by  bending,  this  failure  would  occur 


at  R2.     As  the  bending  moment  at  this 
'  \     |  point  is  1 20  foot-tons  the  section  modulus 


P  is  found  by  multiplying  120  by  ij,  which 

gives  an  I/cof  180.     A  24-inch,  85-pound, 
I-beam  will  be  found  to  have  this  section  modulus. 

For  the  conditions  shown  in  Fig.  45  a  downward  or  negative  reac- 
tion will  occur  at  RI.  Taking  R±  as  a  center  of  moments  and  finding 
RI\  8  tons  x  20  feet  =  160  foot-tons.  One  hundred  and  sixty  foot- 
tons  ~  12  feet  =  13.33  tons  =  R2.  To  find  RI  subtract  R2  from  8 
tons,  or,  8  tons  —  13.33  tons  =  —  5.33  tons  =  RI.  The  minus  sign 
before  5.33  indicates  that  ^i  must  act  in  the  opposite  direction 
from  R%. 


CHAPTER  VIII 

Steel  Framing  Plan.     Record  of  calculations.     "Filling-in"   beams.     Girders. 

Trial  calculations. 

THE  general  principles  outlined  in  the  foregoing  chapters  can 
be  applied  to  nearly  all  conditions  encountered  in  designing 
the  steel  for  a  building.  Just  how  to  apply  such  knowledge,  how- 
ever, is  sometimes  a  question.  There  are  certain  methods  employed 
by  an  engineer  which  are  extremely  simple  in  themselves,  but  which 
are  necessary  if  records  are  to  be  kept  of  the  calculations,  and  which 
also  facilitate  the  work.  The  architect  who  may  know  exactly  how 
to  determine  the  maximum  bending  moment  in  a  beam  and  who 
can  determine  the  radius  of  gyration  of  a  column  section  may  be  at 
a  loss  to  know  how  to  use  these  calculations,  and  a  knowledge  of  the 
methods  referred  to  above  will  undoubtedly  be  of  assistance  to  him. 

The  first  step  to  be  taken  in  determining  the  steel  for  any  build- 
ing, is  to  lay  out  the  framing  plan  for  a  typical  floor.  In  order  to  do 
this  it  is  necessary  to  stretch  tracing  paper  over  the  architect's  plan 
and  locate  the  columns.  The  architect  himself  has  a  great  deal  to 
say  about  the  last  question.  Columns  must  be  placed  so  as  not  to 
interfere  with  the  circulation  and  not  to  project  into  rooms.  In 
loft  buildings,  department  stores,  warehouses  and  buildings  used  for 
manufacturing  purposes,  the  location  of  columns  depends  more 
upon  engineering  than  upon  architectural  requirements.  Archi- 
tects often  want  to  space  columns  too  far  apart  for  engineering  re- 
quirements and  engineers  want  them  too  near  together  to  suit  the 
architect.  If  possible  the  columns  should  divide  the  plan  into  rect- 
angles or  "bays."  These  bays  should  have  approximate  dimensions 
of  eighteen  feet  by  twenty-four  feet.  Of  course  there  are  many 
cases  in  which  such  dimensions  are  impossible  and  the  rectangles  may 
become  larger  or  smaller,  but  a  span  of  over  twenty-four  feet  for 
either  beams  or  girders  gives  excessive  depths  and  a  span  of  less 
than  eighteen  feet  often  results  in  uneconomical  sections. 

Fig.  46  shows  a  portion  of  a  typical  framing  plan  of  a  department 
store.  The  walls  are  self-supporting  and  columns  are  used  only  as 
braces  as  far  as  the  walls  are  concerned.  The  floor  load  is  taken  as 


ENGINEERING  FOR  ARCHITECTS 


63 


200  pounds  per  square  foot.  In  the  bay  included  between  columns 
33,  34,  43  and  44  framing  for  elevator  shafts  and  stair  wells  is  shown. 
Aside  from  this  framing  the  plan  is  perfectly  simple. 

As  shown  the  girders  are  shorter  than  the  beams.     This  is  the 
usual  method  of  framing  because  it  gives  beams  and  girders  of  com- 


FIGURE  46 

paratively  the  same  depth.  If  the  conditions  were  reversed  there 
would  be  excessively  deep  girders  and  beams  that  would  be  — 
architecturally  speaking  —  "out  of  scale." 

In  the  original  layout  the  columns  are  merely  indicated  by  circles 
and  the  beams  framing  directly  to  the  columns  are  drawn  in  as  shown 
in  Fig.  \*j.  At  this  point  it  may  be  noted  that  such  beams  and 
girders  as  shown  must  be  framed  directly  to  the  column  centers  in 


ENGINEERING  FOR  ARCHITECTS 


order  to  secure  stiffness.  Occasionally,  as  between  columns  33  and 
34,  in  order  to  meet  architectural  requirements  it  may  be  necessary 
to  frame  a  girder  a  little  off  center,  but  this  should  never  be  done  in 
such  a  manner  as  to  make  a  direct  connection  between  the  column 
and  girder  impossible. 

Once  this  much  is  drawn,  beams  are  indicated  and  a  tentative 
plan  is  laid  out.     Then  the  actual  work  of  figuring  is  begun. 

In  order  to  have  a  complete  set  of  figures  a  careful  record  of  all 
calculations  must  be  kept.     Some  engineers  use  books,  others  have 

loose  leaf  binders,  and  some  simply  have 
specially  made  pads  of  paper.  Whatever 
the  method,  every  calculation  should  be 
recorded  and  no  figures  which  have  not 
a  direct  bearing  on  the  engineering  prob- 
lems should  be  entered  in  the  record. 

To  start  the  calculations,  a  bay  which 
represents  the  most  common  conditions 
is  chosen.  Such  a  bay  is  found  included 
between  columns  21,  22,  31,  32.  The 

engineer  starts  by  making  a  small  sketch 
FIGURE  47  of  thig  bay  in  hi§  note§  (pig    ^       ^^^ 

having  determined  the  floor  load  he  finds  the  total  uniform  load  which 
a  beam  has  to  carry.  Each  beam  is  24  feet  long  and  carries  a  load 
that  extends  over  an  area  6.6  feet  wide  by  24  feet  long.  The  total 
floor  load  on  the  beam  is  then  31,680  pounds.  Looking  in  the  hand- 
book, the  safe  load  for  an  1 8-inch,  55-pound  I-beam,  spanning  24  feet 
is  39,290  pounds,  so  this  beam  will  be  selected.  This  completes  the 
calculation  for  the  filling-in  beams  and  a  line  is  drawn.  In  connection 
with  the  above  calculation  it  may  be  well  to  call  attention  to  the 
fact  that  although  a  1 5-inch  6o-pound  I-beam  will  carry  the  load, 
the  1 8-inch  beam  will  be  five  pounds  per  foot  lighter  and  this  saving 
for  all  the  beams  in  all  the  floor  panels  will  be  considerable. 

The  next  consideration  is  to  determine  the  size  of  the  girder 
between  columns  31  and  32.  For  the  purpose  a  sketch  is  made 
indicating  the  loads  and  their  positions.  The  loads  are  brought  to 
the  girder  by  the  beams  considered  above  and  act  as  two  concen- 
trated loads.  If  we  consider  that  all  the  weight  of  the  floor  arches 
is  carried  by  the  beams,  there  is  no  uniform  load  on  the  girder.  As 
the  loading  is  symmetrical,  the  reactions  will  be  equal  and  each 
reaction  will  be  31,700  pounds.  The  maximum  bending  moment 


ENGINEERING  FOR  ARCHITECTS  65 

will  be  found  at  the  point  where  either  beam  frames  into  the  girder. 
To  find  I/c  simply  divide  this  bending  moment  in  inch  pounds 
by  S  —  the  safe  working  stress  for  steel  —  or  16,000  pounds  per 
square  inch.  This  is  found  to  be  157  and  a  24-inch  8o-pound  I-beam 
will  be  selected.  The  sizes  of  these  beams  and  girders  must  be 
noted  on  the  plan.  The  girders  and  beams  in  panels  22,  23,  32  and 
33  will  be  the  same  as  those  shown  above. 

In  the  lower  panels  it  may  be  well  to  make  a  trial  to  see  whether 
it  would  be  better  to  frame  the  beams  as  shown  or  to  have  them 
parallel  to  the  longest  dimension  of  the  bay.  First  consider  the 
condition  indicated  on  the  plan. 

A  small  sketch  of  the  bay  should  be  drawn.  The  load  on  the 
beams  will  be  23,800  pounds  and  in  the  handbook  will  be  found  a 
12-inch,  40-pound  I-beam  that  will  carry  the  load  over  a  span  of 
20  feet.  A  12",  2o|-lb.  channel  will  be  strong  enough  to  carry  the 
load  between  columns  41  to  51.  Girder  41-42  will  have  two  con- 
centrated loads  upon  it.  These  loads  will  be  found  by  adding  one- 
half  the  load  on  each  beam  in  panel  21-22-31-32  to  one-half  the  load 
on  each  beam  in  the  panel  at  present  under  consideration.  One- 
half  of  31,680  is  15,840  and  one-half  of  23,800  is  11,900,  so  each  con- 
centrated load  will  be  27,740.  The  section  modulus  will  be  137.2 
and  a  24-inch,  8o-pound  I-beam  will  answer.  To  find  the  size  of 
the  channel  necessary  between  columns  51  and  52,  a  sketch  should 
be  drawn  and  the  two  concentrated  loads  of  11,900  pounds  located. 
The  section  modulus  is  found  to  be  59  and  although  I/c  for  a  1 5-inch 
55-pound  channel  will  be  slightly  less  than  this,  it  will  be  chosen 
as  the  building  department  will  allow  a  slight  difference  of  five 
per  cent. 

The  weight  of  steel  per  square  foot  of  panel  will  be  determined 
as  follows:  One-half  of  the  weight  per  foot  of  the  girder  41-42  or  40 
pounds  is  assumed  to  come  in  the  panel.  The  whole  weight  of  the 
channel  (51-52)  or  55  pounds  is  in  the  panel.  The  weight  of  these 
is  distributed  over  18  feet,  so  to  get  the  weight  per  square  foot  add 
55  to  40  and  divide  by  18.  55  +  40  =  95.  95  •*•  18  =  5.3  pounds 
per  square  foot. 

Each  beam  weighs  40  pounds  per  foot  and  this  is  distributed 
over  an  area  6.6  feet  wide.  So  each  square  foot  would  have 
40  -r-  6.6  =  6  pounds  of  steel  from  the  beams.  Add  6  to  5.3  and  a 
total  weight  of  11.3  pounds  per  square  foot  is  given  as  supplied  by 
beams  and  girders. 


66        ENGINEERING  FOR  ARCHITECTS 

In  order  to  test  the  weight  of  the  panel  with  the  beams  framed 
parallel  to  the  longest  dimension,  a  trial  sheet  should  be  started. 
At  the  top  of  this  the  word  "Trial"  should  be  written.  (Fig.  49.) 
A  sketch  of  the  panel  is  made  with  the  beams  indicated  as  shown. 
The  weight  upon  the  beams  is  then  determined  as  24,000  pounds 
and  a  1 5-inch,  42-pound  I-beam  will  be  selected.  Girder  42-52  will 
have  two  concentrated  loads  of  24,000  pounds  each  and  the  section 
modulus  is  found  to  be  108.  A  2O-inch,  65-pound  I-beam  will 
answer  the  requirements.  For  the  channel  between  columns  41 
and  51  the  section  modulus  of  54  will  be  obtained  —  one-half  of  that 
of  the  girder  —  and  a  1 5-inch,  5O-pound  channel  will  be  strong  enough. 
A  small  sketch  should  be  drawn  for  girder  41-42.  This  carries  a 
uniform  load  and  two  concentrated  loads.  The  total  uniform  load 
can  be  found  —  from  the  calculation  for  beams  in  panels  41,  42,  51 
and  52  —  to  be  12,000  pounds,  and  each  concentrated  load  can  be 
determined  from  the  load  on  the  beams  in  panels  21,  22,  31  and  32. 
These  loads  are  each  15,840.  The  maximum  bending  moment  will 
be  found  at  the  center.  The  upward  moment  around  this  point 
will  be  caused  by  the  reaction  multiplied  by  one-half  the  span. 
The  two  downward  moments  are  caused  by  the  concentrated  load 
(15,840  Ibs.)  and  one-half  the  total  uniform  load. 

The  weight  per  square  foot  of  the  steel  will  be  found  as  before. 
Forty-two  pounds  of  steel  in  the  beam  will  be  distributed  over  6  feet, 
so  there  will  be  7  pounds  of  steel  per  square  foot  from  the  beam. 
50+  32  =  82  pounds  of  channel  and  girder  distributed  over  20  feet 
or  4.1  pounds  per  square  foot.  The  total  weight  will  be  n.i  per 
square  foot  provided  we  assume  that  the  weight  of  girder  41-42  will 
about  counteract  the  light  channel  51-52. 

The  two  weights  will  therefore  be  n.i  in  the  last  panel  and  11.3 
in  the  former,  but  there  is  so  little  difference  in  those  two  weights 
that  it  will  not  be  noticeable.  The  framing  shown  in  Fig.  46  will 
be  adopted  because  a  stiffer  frame  will  result  if  this  method  is  used. 

In  Fig.  50  a  condition  is  shown  where  framing  for  a  stair  well 
occurs  and  where  beam  2-11  carries  a  brick  wall.  The  loading  on 
the  beams  in  panel  10-1 1-20-21  is  the  same  as  that  in  panel  41-42-5 1-52 
or  23,800  pounds.  Then  girder  10-11  carries  two  concentrated  loads 
of  12,000  pounds  approximately  and  a  uniform  load  caused  by  the 
floor.  This  load  equals  10,000  pounds.  The  maximum  bending 
occurs  at  the  center  of  the  beam  and  is  found  to  be  105,400  foot- 
pounds. Now  in  all  previous  calculations  the  usual  method  of 


ENGINEERING  FOR  ARCHITECTS 


67 


determining  I/c  from  this  has  been  to  multiply  by  12  and  divide 
by  16,000.  This  becomes  a  somewhat  lengthy  process  and  the 
same  result  will  be  accomplished  simply  by  multiplying  the  load 
in  thousands  of  pounds  by  f .  The  section  modulus  is  then  found 
to  be  78.6,  and  an  1 8-inch,  55-pound  I-beam  will  be  used.  The  beam 
used  to  support  the  stair  construction  will  carry  a  uniform  load 


24-  x  6.6  *  2fc  '  J/,  6/0 
'F'-Z-tt*"  39,2<?& 


3.5'  20  '/SO'IO.SOO 
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FIGURE  48 

from  one-half  the  floor  arch  and  a  uniform  load  from  one-half  the 
stair  area.  The  weight  per  square  foot  of  stair  construction  is 
taken  as  150  pounds,  including  both  dead  and  live  loads.  The  total 
load  will  be  20,500  pounds  and  a  1 2-inch,  4O-pound  I-beam  will 
support  it. 

Beam  2-n  carries  a  uniform  load  of  the  wall  which  is  13  feet 
high.  Engineers  always  figure  brick  as  weighing  120  pounds  per 
cubic  foot  so  the  wall  will  weigh  1,560  pounds  per  lineal  foot.  The 


68 


ENGINEERING  FOR   ARCHITECTS 


letters  p.  1.  f.  stand  for  the  words  "per  lineal  foot"  in  the  calculations. 
The  reactions  are  found  to  be  13,621  pounds  and  15,329  pounds 
respectively.  The  maximum  bending  occurs  where  the  uniform 


41 


.  _     *f>#         1 

—  I     «-r^i 

'ex     P/?  X 

J>0                            :;J 

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000 


FIGURE  49 

load  is  placed.     The  architect  should  check  this  to  see  if  the  shear 
changes  from  positive  to  negative  at  this  point. 

The  maximum  bending  moment  is  found  to  be  58,395  foot-pounds 
and  the  I/c  is  determined  as  43.7.  A  1 5-inch,  42-pound  I-beam  will 
be  selected.  For  the  channel  i-io  a  section  modulus  of  22  is  ob- 
tained, the  bending  being  caused  simply  be  the  load  brought  to  the 


ENGINEERING  FOR  ARCHITECTS  69 

channel  by  the  beam.  A  1 2-inch,  2oJ-pound  channel  will  be 
selected,  although  it  is  a  little  light.  Channel  1-2  will  carry  only 
such  a  uniform  load  as  is  brought  to  it  by  the  stairs  and  this  will  be 
carried  by  a  1 2-inch,  2o|-pound  channel. 

The  framing  around  the  elevator  shafts  and  stair  well  requires 
but  little  explanation.  Girder  33-34  carries  two  concentrated  loads 
brought  to  it  by  the  floor  beams.  In  some  cases  another  concen- 
trated load  is  considered.  This  is  due  to  the  load  caused  by  the 
impact  of  the  elevator  in  case  the  cables  should  break  and  the 
emergency  brake  cause  the  weight  of  the 
elevator  to  come  upon  the  guides.  In  the 
present  problem  this  load  will  not  be  taken 
into  account  as  engineers  figure  that  the 
live  load  is  great  enough  to  take  care  of 
any  extraordinary  conditions  such  as  this. 
If  the  architect  should  adopt  the  methods 
referred  to  in  this  article  he  will  find  that 
the  girder  required  will  have  to  be  an  18- 

inch,  55-pound  I-beam.  The  stair  well  load  will  be  considered -as  a 
simple  uniform  load  of  150  pounds  per  square  foot  coming  upon  the 
short  beam  and  upon  beam  34-44.  The  short  beam  need  be  only  an 
8-inch,  1 8-pound  I-beam.  The  only  loads  upon  the  beam  framing 
between  beams  33-43  and  34-44  will  be  the  single  concentrated  load 
brought  by  the  8-inch  beam  and  the  weight  of  a  6-inch  terra  cotta 
wall.  A  15-inch,  25-pound  I-beam  will  be  found  to  be  strong 
enough.  A  lo-inch,  5<D-pound  I-beam  will  be  framed  between  columns 
33  and  43  and  a  lighter,  42-pound  beam  between  34  and  44. 

The  methods  given  in  this  article  are  mere  suggestions  of  those 
that  may  be  employed  in  actual  practice.  The  conditions  that  have 
been  considered  are  typical  ones  and  although  there  may  be  many 
more  complicated  ones,  the  architect  need  have  no  fear  of  them  pro- 
vided he  is  careful  in  his  diagnosis  of  the  case.  The  chief  diffi- 
culties encountered  are  not  due  to  engineering  considerations,  but 
to  carelessness  in  placing  the  loads  or  in  leaving  important  loads 
out  of  the  calculations  entirely.  The  architect  is  too  apt  to  want  to 
start  in  figuring  before  he  fully  draws  out  the  diagram  that  repre- 
sents the  methods  of  loading  the  beam. 

Architects  often  believe  that  engineers  have  a  great  guessing 
ability  and  do  not  figure  out  all  the  conditions.  This  impression  is 
absolutely  wrong.  Records  of  the  best  engineers  in  the  profession 


70        ENGINEERING  FOR  ARCHITECTS 

show  that  every  little  detail  is  gone  into,  that  there  is  not  a  single 
beam  in  a  floor  plan  that  has  not  been  very  carefully  considered. 
Those  who  are  most  familiar  with  the  situation  realize  that  even 
with  the  most  careful  checking  mistakes  will  occur  and,  because  of 
this,  as  little  as  possible  is  left  to  chance. 

An  architect  who  found  that  certain  considerations  had  been 
overlooked  in  his  plans,  excused  his  oversight  to  the  engineers  by 
saying  that  he  had  taken  certain  things  for  granted.  The  answer 
was  made,  "Be  a  good  engineer  and  take  nothing  for  granted." 


CHAPTER   IX 

Column  Schedules.     Formulas.    Loads.    Column  sections.    Checks.    Eccentric  loads. 

THE  sets  of  steel  drawings  that  an  architect  is  usually  most 
familiar  with  are  the  framing  plans  and  the  column  schedule. 
The  method  of  determining  the  sizes  of  beams  in  a  framing  plan 
was  taken  up  —  more  or  less  roughly  —  in  the  last  chapter.  This 
chapter  will  deal  with  the  considerations  involved  in  the  working  out 
of  a  column  schedule.  Fig.  51  shows  a  portion  of  such  a  schedule. 
For  the  purposes  of  this  chapter  all  the  columns  are  made  of  plates 
and  channels,  because  tables  are  given  for  the  safe  loads  to  be  car- 
ried by  such  columns  in  both  the  Cambria  and  Carnegie  handbooks. 

In  order  to  understand  the  methods  employed  by  engineers  in 
developing  column  schedules,  the  architect  should  become  thor- 
oughly familiar  with  the  formulas  given  in  Chapter  V.  The  stress 
per  square  inch  that  is  allowed  by  the  New  York  Building  Depart- 
ment is  given  by  the  formulas  S  =  15, 200  — 58  //r,  /  being  the 
unsupported  length  of  the  column  in  inches  and  r  the  radius  of 
gyration  of  the  column  section — also  measured  in  inches.  The 
method  of  determining  r  should  also  be  remembered  as  there  often 
exist  cases  where  ordinary  column  sections  given  in  the  handbook 
cannot  be  used. 

To  these  formulas  another  should  be  added — -one  dealing  with 

W  -\-W       Me 

the  conditions  of  eccentric  loading.     This  is  S=—          - -\ — r.     In 

A  1 

this  formula  it  is  assumed  that  two  kinds  of  loads  exist,  one,  W\y 
coming  directly  upon  the  axis  of  the  column  and  the  other,  /F2, 
acting  at  a  distance  more  than  eight  inches  away  from  the  center 
line.  Wi  is  the  axial  or  concentric  load  and  W2  is  the  eccentric  load. 
If  no  attention  is  to  be  paid  to  the  bending  set  up  in  a  column 
due  to  the  eccentric  load,  it  is  plain  that  S  the  stress  per  square 

W  -f-  W 

inch   due   to   W^  and   W^  will   be  -    — -. — -  in  which   formula  A   is 

A 

the  area  of  the  cross  section.  This  is  the  first  part  of  the  for- 
mula given  above.  There  will  be,  however,  extra  bending  stresses 
due  to  the  eccentric  load  —  W2.  In  Fig.  52  it  will  be  seen  that 
the  load  W2  is  acting  at  a  distance  y  from  the  center  of  the  column. 


72        ENGINEERING  FOR  ARCHITECTS 

The   moment  set   up   by   this   load   will    be    Wy   and    this  will   be 
denoted  by  M. 

If  the  architect  remembers  the  formula  for  the  resisting  moment 
of  a  beam,  M  =  S  I/c,  he  can  easily  see  that  S  =  M  -f-  I/c.  If  this 

W  4-  W 

S  is   added   to   that   caused  by  the  concentric  load  -    ~ — -  then 

A 

the  total  S  in  the  section  will  be  determined.     So  the  whole  for- 

•11   i      r.      W\  H~  W*  ,  Me.     ^, 

mula  will   be  S=—  -^ — -  +  — r~     The   use   made   of   this   formula 
A  1 

will  be  referred  to  later. 

The  first  thing  to  be  determined  is  the  load  upon  the  column. 
In  order  to  do  this  and  also  keep  a  fairly  accurate  record  of  all 
his  calculations,  the  architect  should  adopt  some  form  of  tabu- 
lating his  loads.  The  following  method  of  recording  loads  is  sub- 
mitted, not  as  an  absolute  and  set  form,  but  as  one  that  has 
proved  fairly  satisfactory  and  one  that  will  furnish  the  architect 
with  a  means  of  attacking  the  ordinary  problems  involved  in 
column  design. 

Fig.  53  shows  a  sheet,  ruled  in  such  a  manner  as  to  allow  for 
the  tabulation  of  the  loading  upon  the  columns  on  each  floor.  The 
reason  for  dividing  the  live  load  from  the  dead  loads  is  that  the 
building  department  allows  a  reduction  of  five  per  cent,  in  the  live 
load  for  each  floor  below  the  roof  and  top  floor  until  fifty  per  cent, 
of  the  load  is  deducted  and  after  that  no  further  deduction  is  al- 
lowed. A  space  is  left  for  added  loads  due  to  eccentric  loading. 

To  explain  the  method  of  proceeding  an  example  will  be  given. 
Column  21,  as  seen  in  Figure  54,  is  an  interior  column  having  beams 
and  girders  framing  to  it  in  the  ordinary  manner.  There  is  no  ec- 
centric loading  at  all.  Let  it  be  considered  that  the  live  load  is  120 
pounds  per  square  foot  and  the  dead  load  80  pounds  per  square  foot, 
the  total  being  200  pounds,  60  per  cent,  of  which  is  live  and  40  per 
cent,  dead  load. 

The  quickest  method  of  finding  the  load  on  columns  is  to  con- 
sider the  area  of  floor  which  the  column  carries.  Such  an  area  is 
included  within  the  lines  ab,  be,  cd,  and  da  (Fig.  54)  and  is  19  feet 
wide  by  24  feet  long.  The  weight  of  this  is  24  X  19  X  200  =  91,200 
pounds,  54,720  pounds  being  live  load  and  36,480  pounds  dead  load. 
A  record  of  these  calculations  is  entered  on  the  ruled  sheet,  Fig.  53, 
the  live  load,  dead  load,  and  total  all  being  shown,  as  coming  upon 
the  column  at  the  eighth  floor. 


ENGINEERING  FOR  ARCHITECTS 


73 


The  roof  load  differs  from  that  of  the  eighth 
floor,  in  that  the  live  load  in  the  roof  is  only  50 
pounds  as  required  by  the  Building  Code  and 
the  total  load  will  be  80+  50  =  130  pounds  per 
square  foot.  The  load  upon  the  column  at  the 
roof  level  will  be  24  X  19  X  130  =59,300  pounds  — 
22,800  pounds  live  and  36,500  pounds  dead  load. 

The  load  at  the  seventh  floor  would  be 
exactly  the  same  as  that  on  the  eighth  except 
that  the  live  load  is  reduced  five  per  cent.  In 
this  connection  it  may  be  well  to  say  that  the 
process  of  dropping  off  this  percentage  of  the 
live  load  at  each  floor  may  be  made  very  simple 
by  the  use  of  the  slide  rule.  Dead  loads  are 
simply  recorded  as  shown.  Ninety-five  per  cent, 
of  54,720  is  approximately  52,000  pounds. 
Ninety  per  cent,  is  49,300  pounds  and  so  on. 
The  total  load  brought  to  the  column  by  each 
floor  is  then  found  and  these  totals  are  added 
together. 

The  next  step  is  the  selection  of  the  column 
sections  that  will  withstand  these  loads.  It 
will  be  noted  on  the  schedule  that  the  columns 
are  made  in  sections  extending  through  two 
floors  —  that  one  section  extends  from  the 
basement  to  the  second  floor,  and  another 
from  the  second  to  the  fourth  floor.  The  last 
section  will  extend  from  the  sixth  floor  to  the 
roof  and  the  greatest  load  upon  it  is  238,980 
pounds.  Looking  in  the  table  for  safe  loads 
for  columns  it  will  be  found  that  for  an  unsup- 
ported length  of  12  feet  a  section  made  of  10- 
inch,  15-pound  channels  and  15  by  f-inch  plates 
will  support  245,000  pounds.  This  must  be 
checked  to  see  if  it  conforms  to  the  Building 
Code  requirements,  r  =  4.65,  /  =  12x12=  144. 
58  (/  +  r)  =  58  X  144  *  4.65  =  1,795.  15.200  - 
1,795  =  13,405  =  S. 

The  area  of  the  section  is  20.17,  so  the 
load  it  will  support  is  20.17  X  13,405  =  270,000 


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74 


ENGINEERING  FOR  ARCHITECTS 


pounds.  A  lighter  column  might  be  used  so  the  section  will  be  made 
of  the  channels  with  5/i6-inch  plates.  It  might  be  well  to  state 
that  the  Carnegie  formula  for  stresses  in  columns  is  different  from 
that  employed  in  determining  the  safe  loads  for  Cambria  column 
sections  and  both  differ  from  the  formula  employed  by  the  New 
•yW*  \AA  York  Building  Department,  but  the  results 

^  are  about  the  same  in  all  cases.  The  portion 
J^  of  column  extending  from  the  seventh  floor 
to  the  roof  is  much  heavier  than  necessary, 
and  in  some  cases  engineers  take  this  into 
account,  but  the  saving  resulting  from  the 
cutting  down  of  field  riveting  and  splicing 
and  the  extra  stiffness  of  the  frame  makes 
the  use  of  a  long  section  of  column  advan- 
tageous. The  next  section  extends  from  the 
fourth  to  the  sixth  floor  and  the  load  of 
407,840  pounds  at  the  fifth  floor  determines 
the  design. 

The  section  given  in  the  handbook  as 
being  strong  enough  to  withstand  this  load 
is  made  of  1 2-inch  channels  weighing  30 
pounds  per  foot,  and  1 6-inch  by  7/1 6-inch 
plates.  The  radius  of  gyration  of  this 
section  is  5.04  and  58  X  144  •*•  5.04  =  1,653. 
Subtracting  this  from  15,200,  the  stress  per 
square  inch  is  given  as  13,547  pounds.  The 
area  of  the  section  being  31.64  square  inches, 
the  total  strength  of  the  section  will  be  31.64 
This  section  is  safe  and  will  be  used.  To 
jump  from  a  lo-inch  channel  section  to  a  1 2-inch  section  requires 
the  use  of  a  butt  plate  and  angles.  Such  a  connection  is  shown 
on  page  308  in  the  Cambria  handbook,  and  on  page  279  of  the 
Pocket  Companion  published  by  the  Carnegie  Steel  Company, 
and  is  indicated  by  the  use  of  double  lines  on  the  column 
schedule. 

The  section  of  column  21  extending  from  the  fourth  to  the  second 
floor  must  support  a  load  of  565,700  pounds.  It  will  be  found  that 
a  section  made  up  of  1 5-inch,  4O-pound  channels  and  1 7-inch  by 
9/i6-inch  plates  will  be  strong  enough. 

The  last  section  will  have  to  support  a  load  of  712,610  pounds 


FIGURE  52 
X  13,547  =    428>6oo. 


ENGINEERING  FOR  ARCHITECTS  75 

at  the  first  floor  and  a  section  made  of  15-inch,  55-pound  channels 
and  i y-inch  by  f-inch  plates  will  be  strong  enough. 

The  next  column,  number  22,  extends  to  the  pent  house  and 
carries  the  load  of  elevator  beams  and  a  portion  of  the  pent  house 
roof.  In  calculating  the  sizes  of  beams  necessary  to  carry  the 
elevator  sheave  beams  the  designer  had  to  determine  the  reactions 
brought  to  the  columns  by  these  beams.  To  find  the  total  load 
upon  the  column  all  that  is  necessary  to  do  is  to  add  the  proper 
reactions  together.  For  the  design  of  columns  alone  it  is  absolutely 
necessary  for  the  engineer  or  architect  to  keep  complete  records 
of  all  the  calculations  for  beams  in  the  framing  plans. 

There  is  no  separation  of  live  and  dead  loads  for  the  roof  and 
pent  house  beams  as  there  is  no  percentage  of  the  live  load  deducted 
until  the  eighth  floor  is  reached.  It  will  be  noticed  on  the  column 
schedule  that  a  section  extends  from  the  sixth  to  the  eighth  floor 
and  another  from  the  eighth  to  the  pent  house  roof.  Both  for  the 
economy  of  material  in  the  column  section  and  for  the  requirements 
of  shipping  this  is  necessary.  The  load  that  the  top  section  must 
stand  is  found  at  the  roof.  This  is  206,435  pounds,  and  a  section 
made  of  two  lo-inch,  1 5-pound  channels  and  two  1 5-inch  by  J-inch 
plates  will  answer.  The  rest  of  the  calculations  are  the  same  as 
for  column  21. 

So  far  all  loads  have  been  concentric  or  axial  loads,  but  column 
23  has  to  support  a  wall  load  which  sets  up  bending  stresses  in  the 
column.  The  method  of  finding  column  sections  of  sufficient 
strength  to  withstand  eccentric  loads  is  not  direct,  but  involves 
assuming  sizes  and  then  checking.  Of  the  55,970  pounds  brought 
to  column  23,  25,500  pounds  are  eccentric  as  they  are  due  to  wall 
loads.  The  beams  that  carry  the  wall  are  8  inches  ofF  the  center. 
(Fig.  54.)  The  moment,  M ,  set  up  by  the  eccentric  load  is  25,500  X 
8  =  204,000  inch-pounds. 

Referring  to  the  formula  in  the  first  part  of  this  article,  it  will  be 
found  that  the  added  stress  per  square  inch  due  to  the  eccentric  load 
is  M  -7-  I/c  or  Me  1 1.  I  /  c  for  a  column  section  such  as  those  used  in 
this  article  can  be  found  in  the  handbook  under  the  heading  "Mo- 
ments of  Inertia  and  Section  Moduli  for  Plate  and  Channel  Col- 
umns." Assuming,  for  reasons  that  will  be  given  later,  that  a 
section  made  of  two  1 2-inch,  20. 5-pound  channels  and  two  plates 
16  inch  by  7/16  inch  will  be  strong  enough,  I/c  can  be  found  to  be 
123.9.  M  +  I/c  will  be  20,400-7-  123.9=  l>£>47-  This  is  the  stress 


ENGINEERING  FOR  ARCHITECTS 


Col    #ZO 

Col.*&/ 

Co/*Z2 

s~*         r    ^o  t? 
C  0  /     /O  vJ 

1 

1 

Vead 

\ 

\ 

Live, 

4v 

)    UA 

Lcc. 

#\- 

V  jr 

Total 

t 

p 

105,  (>00 

Section 

-  ,    J 

X 

Veact 

36,500 

Lil/e 

ZZ.  300 

55^70 

FLcc. 

4Z.400 

ToM 

53.  300 

100,835 

206,435 

Sect* 

Z  10"  111  -15 

f2$l5t;j. 

* 

Vead 

3^48^ 

25  600 

43,  703 

Liu  & 

$4,720 

38,400 

27,835 

£Tcc. 

53.  DM 

Total 

yi,  zoo 

150,500 

6>4,  000 

Z70A35 

11,  600 

IZ7.570 

Sect'n 

I 

Vecic/ 

36,480 

25,600 

43,  705 

Lwe 

5Z.OOQ 

36.480 

26.400 

Ecc. 

59,  000 

Total 

238,980 

62,  080 

332,5/5 

70,105 

mt>75 

Seci'n. 

Z-/0"&ti 

*2&l5t% 

2  JO"M/5* 

££>hl5"rft 

2Y2"^/,22 

[Zpk  !£>'<•  4. 

1 

va 

Vend 

36.4SO 

25  60) 

43,705 

Live 

45,300 

34,600 

25,060 

Lcc. 

5Z.  000 

Total 

85,  7&0 

324,760 

60,200 

332,7/5 

68,  765 

Zbt>,44l 

tec  fri. 

1 

Pead 

36,4QC 

Z5,  600 

43J05 

Live. 

46,600 

32.600 

23,  700 

Lcc. 

5Z.  000 

lota/ 

8>3,0&) 

407.  840 

58,  ZOO 

4503/5 

&7AQ5 

333,845 

Z>c,ot*n. 

1-IZ'U3(? 

2W/6*/l 

'Z-IZ'M30 

Z/y/s/ti'k 

2/5  HI  53' 

Zphlli  \ 

\ 
$ 

Tfeacl 

3£>,480 

25,600 

43,  705 

Lil/e 

43,800 

30,  700 

22,300 

Lcc 

52,  000 

Total 

60.280 

4&8./20 

56,  30C 

507.215 

66,005 

3W.850 

Secfn 

| 

Veact 

36,480 

Z5,600 

43.705 

Live 

41.  100 

2&.800 

ZO  .  900 

Tec 

5Z,000 

Total 

17,580 

5l>5,  700 

54-,4-OL 

561.  615 

£>4,  60^ 

464.45$ 

Sccfn 

Z-l5"m4tf 

Zfyk/7*fc 

Z'  l5'$14L 

21*1*17'%, 

Z/yk/T'K, 

i 

Peaof 

36,480 

Z5.&M 

A3.  705 

Live 

35,350 

Z6.  350 

19.300 

L.CC 

5Z.  000 

Total 

74.830 

M4.530 

52,450 

&4.065 

£>3.t05 

5  £7.  666 

1 

T?ead 

^\ 

364BD 
35  6DO 

Z5,  600 

4-3,705 

Live- 

I  ffl 

Z4.95C 

I&.IOO 

Ecc. 

v  (\\Y 

K        V 

52  ODO 

Total 

BJE 

ss 

7Z,  080 

7l£,  t>IO 

51,550 

665,  6/5 

6ZJ3G* 

589,4(£ 

beoi'n 

s 

1  15'  ft  55 

2^174 

2-15  '$55 

2  ^17  4 

TWJ& 

'fyblhk 

FIGURE  53 


ENGINEERING  FOR  ARCHITECTS 


77 


per  square  inch  set  up  by  the  eccentric  load.  To  find  the  load  that 
would  cause  an  equal  stress  per  square  inch  if  exerted  over  the  entire 
area  of  the  column,  simply  multiply  this  unit  stress  by  the  area  of 
the  section.  1,647x26.06=42,900  pounds.  This  load  is  placed 
in  the  outside  column  in  Fig.  53  and  is  not  added  until  the  area  of 
the  section  of  the  column  is  to  be  determined. 

The  same  process  is  gone  through  to  find  the  loads  for  the  eighth 
floor.  The  actual  eccentric  load  being  35,000  pounds,  it  will  be 
found  that  a  concentric  load  of  59,000  pounds  will  have  the  same 
effect  as  the  moment  set  up  by  the  eccentricity.  It  will  be  noted 
that  there  is  no  separation  of  the  live  and  dead  loads  at  the  roof 
level,  but  that  this  must  be  done  at  the  eighth  floor.  The  same 


f^ 


© 


4 


© 


@l 


<§> 


-<5" 


FIGURE  54 

process  is  gone  through  for  the  seventh  floor,  but  the  live  load  is 
reduced  5  per  cent. 

The  section  of  column  extending  from  the  sixth  floor  to  the  roof 
will  have  to  be  designed  to  carry  the  load  of  256,675  pounds  —  the 
sum  of  the  load  brought  to  the  column  by  the  roof,  the  eighth,  and 
seventh  floors  plus  the  greatest  load  due  to  eccentricity.  In  the 
table  of  safe  loads  a  column  made  of  1 2-inch,  2oJ-pound  channels 
and  1 6-inch  by  J-inch  plates  will  give  about  the  proper  strength. 


The    radius    of   gyration    is    5.23.      15,200-  58  X   -     -  =  15,200- 

1,600  =  13,600  pounds   per  square  inch.     Multiplying  this   by   the 
area  and  the  result  —  272,000  pounds  —  will  be  near  enough. 

The  next  section  will  only  have  to  be  made  strong  enough  to 
withstand  the  concentric  load  of  197,675  pounds  brought  down  by 
the  upper  section,  plus  the  floor  loads  at  the  sixth  and  fifth  floors 


78  ENGINEERING  FOR  ARCHITECTS 

and  the  eccentric  load  at  one  floor.  The  eccentric  loads  on  the  top 
section  will  not  be  added  to  those  of  the  lower  portion.  Thus  it 
may  be  assumed  that  there  will  be  an  added  concentric  load  of 
about  59,000  pounds  approximately.  So  at  the  fifth  floor  level 
there  will  be  an  assumed  load  of  197,675+  140,000+59,000  = 

396>675- 

For  this  load  it  will  be  better  to  choose  a  section  made  of  1 5-inch 
channels,  as  the  weight  of  a  light  1 5-inch  channel  section  will  be  about 
the  same  as  a  fairly  heavy  1 2-inch  channel  column,  and  the  section 
modulus  will  be  much  larger,  giving  more  resistance  to  bending. 
The  section  modulus  of  a  column  made  of  1 5-inch,  33-pound  channels 
and  17-inch  by  f-inch  plates  is  175.1  and  the  area  32.55  square 
inches.  The  added  eccentric  load  will  be  52,000  pounds.  Adding 
the  loads,  as  shown  in  Fig.  53,  the  load  at  the  fifth  floor  will  be 
333,845  +  52,000  =  385,845  pounds.  Checking  the  section  as  be- 
fore it  will  be  found  to  be  strong  enough  to  withstand  this  load. 

Applying  the  principles  outlined  in  this  chapter  to  the  remain- 
ing sections  the  results  will  be  found  to  correspond  to  the  figures 
in  the  column  schedule. 


CHAPTER  X 


Graphical  Methods  of   Indicating  Forces.     Composition  of  forces.     Resultants  and 
equilibrants.     Points  of  application.     Graphical  methods  applied  to  simple  beams. 

A  FORCE  is  completely  defined  when  its  magnitude,  direction, 
and  point  of  application  are  known.  In  Fig.  55  is  a  simple 
beam  with  a  concentrated  load  upon  it.  The*  magnitude  of  the 
load  is  given  by  the  figures,  —  2,000  pounds,  —  the  direction  is 
indicated  by  the  arrows,  —  downward  —  and  the  location  by  the 
dimensions.  In  this  manner  the  load  is  entirely  defined.  There 
is,  however,  another  method  of  showing  the  three  elements  of  a 
force  and  that  is  by  means  of  a  straight 
line.  In  Fig.  55^  the  line  is  shown  M'-*'  2ffi°  6-a- 
divided  into  equal  parts.  Each  part  is 
supposed  to  represent  1,000  pounds; 
the  total  length  of  the  line  represents 
2,000  pounds.  In  this  manner  the  length 
of  the  line  represents  the  magnitude  of  the  force. 


I 


FIGURE  55 

The  arrow  shows 
the  direction,  and  the  point  p  may  give  the  point  of  application. 
As  lines  may  represent  forces,  it  is  possible  to  use  them  to  deter- 
mine the  action  of  forces.  Two  forces  acting  at  a  point  may  be 
replaced  by  a  single  force  acting  at  that  point.  In  Fig.  56  two 
forces,  PI  and  P2>  act  at  the  point  p.  The  same  effect  will  be  pro- 
duced if  a  single  load  —  R  —  acts  at  that  point.  The  method  of 
determining  R  is  by  means  of  the  "parallelogram  of  forces."  It 

will  be  noticed  that  R  is  the  diagonal  of 
a  parallelogram  the  sides  of  which  are 
parallel  to  PI  and  P2  respectively.  Instead 
of  constructing  a  parallelogram  a  triangle 
may  serve  the  same  purpose  (Fig.  $6a)  in 
which  PI  and  P2  form  two  sides  and  R 
the  third.  R  is  always  used  to  designate 
the  single  force  which  will  take  the  place  of  a  system  of  several 
forces  and  is  called  the  resultant.  To  find  the  magnitude  and  direc- 
tion of  the  resultant  of  two  forces,  all  that  is  necessary  to  do  is  to 
construct  a  triangle,  as  shown  in  Fig.  560,  with  the  two  known 


8o 


ENGINEERING  FOR  ARCHITECTS 


FIGURE  57 


forces  represented  parallel  to  their  original  direction  and  the  arrow 
head  of  one  placed  at  the  butt  of  the  other. 

Fig.  57  represents  another  condition.  In  this  case  it  is  neces- 
sary to  find  not  the  resultant,  but  a  force  that  will  oppose  the  two 
known  forces  and  set  up  a  condition  of  equilibrium.  E  is  found  in 

the  same  manner  as  jR,  but  it  will  be  seen 
that  the  arrow  points  in  an  opposite  direc- 
'•  tion.  This  force,  if  acting  at  p  at  the 
same  time  as  PI  and  P2,  will  force  p  to 
remain  exactly  as  it  is.  E  then  sets  up  a 
condition  of  equilibrium  and  is  known  as 
the  equilibrant.  The  equilibrant  is  equal 
in  magnitude  to  the  resultant,  but  acts  in  the  opposite  direction. 
In  Fig.  57&  the  forces  are  shown  in  the  same  manner  as  in  56^,  but 
it  will  be  noticed  that  all  the  arrows  tend  to  follow  around  the 
triangle.  When  this  is  the  case  there  is  always  a  condition  of 
static  equilibrium. 

So  far  there  have  only  been  two  known  forces,  but  it  is  possible  to 
determine  the  magnitude  and  direction  of  the  resultant  or  equilib- 
rant of  any  number  of  forces  provided  the  magnitude  and  direction 
of  each  one  of  them  is  known.  In  Fig.  58  are  shown  four  forces 
PI,  P2,  P3,  and  P4  acting  in  the  directions  indicated. 

PI  and  P2  can  be  drawn  as  shown  and  resolved  into  the  resultant 
RI.  A  second  triangle  made  of  RI  and  P3  and  R2  can  be  formed, 


FIGURE  58 

RZ  being  the  resultant  of  RI  and  P3.  A  third  triangle  made  of  R2y 
P4,  and  R3  will  give  the  resultant  of  the  two  forces  R%  and  P4.  From 
this  it  can  be  seen  that  R3  is  the  resultant  of  all  four  forces,  PI,  P2, 
P3  and  P4.  Instead  of  drawing  RI  and  R2  in  order  to  find  the  resultant 
of  a  system  of  forces,  all  that  is  usually  drawn  is  the  final  resultant 
as  shown  in  the  third  diagram  of  Fig.  58.  It  will  be  noticed  that  all 


ENGINEERING  FOR   ARCHITECTS  81 

the  arrows  except  that  of  R  follow  around  the  diagrams.  To  find  the 
equilibrant  of  three  forces  the  same  method  is  employed  as  shown 
in  Fig.  59.  Note  the  direction  of  the  arrows. 

Although  this  method  gives  the  magnitude  and  direction  of 
either  R  or  E  the  points  at  which  such  forces  are  to  be  applied  —  so 
as  to  give  the  same  results  as  the  system 
of  forces  or  to  put  the  system  into  a  con-  v 
dition  of  equilibrium  —  has  not  yet  been  \  pi 
determined.  A  further  elaboration  of  *i  \ 
the  figure  will  give  the  required  points. 

In    Fig.    56   it   was   shown   that   two 

forces    could    be   resolved   into    a    single 

r  r>  •         i_  •  FIGURE  59 

force,     r>y  reversing  the  process  it  can  be 

shown  that  a  single  force  may  be  divided  into  two  forces.  PI  in 
Fig.  6ob  acts  through  the  point  p.  The  magnitude  of  PI  is  shown 
by  the  line  AB  (Fig.  6oa).  To  get  the  magnitude  and  direction  of 
any  two  forces  that  will  give  the  same  result  as  PI,  take  any  point 
0,  and  draw  OA  and  OB.  Also  draw  0' A'  and  O'B'  in  Fig.  6ob 
parallel  to  OA  and  OB  and  passing  through  p.  These  two  forces 
will  produce  the  same  effect  on  p  as  the  single  force  PI.  To  show 
that  0  can  be  taken  at  any  point,  note  the  effect  in  Fig.  60 
(c  and  d). 

Fig.  61  shows  a  system  of  four  forces  divided  up  in  this  manner. 
OA,  OB,  OC,  OD  and  OE  are  the  components  of  Pl9  P2,  P3  and  P4, 
respectively.  These  can  be  transferred  to  A  as  shown  starting  with 

any  point  p  on  PI  and  drawing  O'A', 
O'B',  O'C',  etc.,  parallel  to  OA,  OB, 
OC,  etc. 

It  will  be  noticed  that  OA  and  OE 
act  as  components  of  R  in  the  same 
manner  as  OA  and  OB  act  as  compo- 
nentsofPi.  At  the  intersection  of  O'A' 
and  O'B',  (Fig.  6ia)  a  point  is  found 
through  which  R  must  act.  In  this 

manner  the  point  of  application  of  R  can  be  determined,  and  all 
that  it  is  necessary  to  know  about  the  resultant  is  known  —  the 
magnitude  and  direction  being  given  in  b. 

At  this  point  it  is  suggested  that  the  architect  draw  several 
diagrams  of  his  own,  selecting  any  number  of  forces  of  any  magni- 
tude and  find  the  resultants  and  points  of  application. 


82 


ENGINEERING  FOR  ARCHITECTS 


FIGURE  61 


By  using  the  same  method,  it  is  possible  to  find  the  direction 
and  magnitude  of  two  unknown  parallel  forces,  provided  we  know 
their  points  of  application.  Given  the  forces  PI,  P2,  P3  and  P4,  as 
shown  in  Fig.  62,  and  the  two  points  X  and  T,  through  which  it 

is  desired  to  pass  two  forces,  parallel  to 
each  other,  that  it  will  set  up  a  condition 
of  equilibrium.  In  b  the  diagram  — • 
known  as  the  fo rcei polygon  —  is  drawn, 
in  which  PI,  P2,  PS  and  P4  are  carefully 
laid  out  equal  in  magnitude  (length)  and 
parallel  in  direction  to  the  respective 
forces  in  a.  F  and  A  are  connected  and 
the  direction  of  the  two  parallel  forces  is  determined.  Through  X 
and  T  draw  lines  parallel  to  FA.  It  is  now  necessary  to  find  the 
magnitude  of  these  two  forces.  Locate  0  at  any  point  and  draw 
OA,  OB,  etc.  Through  any  point  on  PI,  draw  0 'A'  and  O'B'  and 
complete  the  diagram.  It  will  be  noticed  that  continuing  0' A'  until 
it  intersects  E\,  and  also  continuing  O'F'  until  it  intersects  E%, 
points,  on  EI  and  E2  are  found  that  must  be  connected  by  O'J'  to 
close  the  diagram.  By  drawing  OJ  parallel  to  O'J'  from  0  to  a  point 
on  FA,  FA  is  divided  into  two  parts.  FJ  gives  the  magnitude  of 
E2  and  JA  gives  the  magnitude  of  E\.  In  this  manner  it  is  possible 
to  determine  the  magnitude  of  two  parallel  forces,  acting  through 
any  two  points  that  will  set  a  system  of  forces  in  equilibrium. 


FIGURE  62 

Now  the  obvious  question  is,  what  is  the  use  of  all  this?  Per- 
haps the  answer  is  best  found  in  a  practical  problem.  Given  a  beam 
1 8  feet  long  and  carrying  loads  shown  in  Fig,  63,  it  is  necessary  to 
find  the  reactions.  Lay  off  the  polygon  as  indicated  in  (b)  with  AB 
parallel  in  direction  and  equal  in  magnitude  to  the  120  pound  load, 


ENGINEERING  FOR   ARCHITECTS  83 

BC,  and  CD  bearing  the  same  relations  to  the  580  and  600  pound 
loads,  respectively,  and  the  sum  —  AD  —  representing  the  total 
load  upon  the  beam. 

Complete  the  diagram  in  (a)  and  it  will  be  found  that  O'A' 
intersects  RI  and  O'D'  intersects  R2  and  that  these  intersections 
may  be  connected  by  O'F'.  Drawing  OF  in  (b)  parallel  to  O'F', 
DA  is  divided  into  two  forces,  DF  which  is  equal  to  R2,  and  FA', 
which  is  equal  to  RI.  Scaling  off  these  lengths,  it  can  be  found 
that  RI  equals  380  pounds,  and  R2  equals  920  pounds.  These  results 
can  be  checked  as  follows: 

120  pounds  X    5  feet  =      600  foot-pounds 
580  pounds  X  12  feet  =  6,960  foot-pounds 
600  pounds  X  15  feet  =    9,000  foot-pounds 
Totals  1,300  pounds  16,560  foot-pounds 

16,560-=-    18  =  920 
1,300  -  920  =  380 

By  combining  (a)  and  (b)  in  diagram  (c)  the  shear  diagram  is 
easily  drawn.  The  base  line  is  drawn  in  line  with  F  (b).  The 
shear  at  R1  is  drawn  parallel  and  equal  to  FA.  The  loads  "step"  off 
as  shown,  all  points  being  transferred  from  either  a  or  b.  It  will  be 
found  that  this  is  a  graphical  method  of  showing  that  the  sum  of 
the  reactions  must  equal  the  sum  of  the  loads.  Another  thing 
worthy  of  notice  is  that  the  forces  AB,  BC  and  CD  are  read  down- 
ward and  DF  and  FA  are  read  upward,  and  that  both  diagrams  are 
closed.  This  means  that  R\  and  R2  are  upward  forces  and  that  for 
conditions  of  equilibrium  the  diagrams  must  close.  The  point  of 
no  shear  is  found  under  the  58o-pound  load  and  this  might  be  told 
from  the  fact  that  F  falls  between  B  and  C. 

So  far  we  have  been  able  to  find  the  reactions  and  the  shear 
diagram  by  means  of  the  graphical  methods.  It  can  be  shown 
that  the  bending  moment  at  any  point  in  the  beam  can  also  be  deter- 
mined. Let  it  be  necessary  to  find  the  maximum  bending  moment. 
This  occurs  under  the  58o-pound  load. 

Two  new  lines  must  be  drawn,  a  perpendicular  is  dropped  from 
0  to  AD  and  the  length  of  this  line  is  designated  as  H.  This  is  known 
as  the  pole  distance,  and  0  is  known  as  the  pole.  In  other  words  the 
perpendicular  distance  from  the  pole  to  the  resultant  is  the  pole 
distance.  This  is  measured  in  the  same  units  as  AB  or  BC,  or 


84 


ENGINEERING  FOR   ARCHITECTS 


force  units.  As  0  can  be  selected  at  any  point,  the  distance  H  can 
be  taken  as  any  distance,  but  for  convenience  it  is  usually  given 
some  even  value.  In  this  case  it  is  equal  to  a  distance  that  would 
scale  1,000  pounds. 

The  other  line  is  one  parallel  to  AD  and  passing  through  the 
58o-pound  load,  cutting  the  diagram  at  /  and  g.     As  all   distances 

in  (a)  are  measured  in  feet  the 
distance  fg  is  found  to  scale 
3.72  feet. 

To  find  M  under  the  580 
s*  pounds  all  that  it  is  neces- 
sary to  do  is  to  multiply 
H  (in  pounds)  by  fg  (in 
feet)  and  the  product,  3,720 
foot-pounds,  will  be  the  max- 


FIGURE  63 

imum  bending  moment.  In  this  manner,  all  the  information  neces- 
sary for  the  design  of  a  beam  can  be  found  by  graphical  methods. 
Checking  the  value  for  M: 

+  380  pounds  X  12  feet  =  4,560  foot-pounds 

-  120  pounds  x    7  feet  =      840  foot-pounds 

M  =  3,720  foot-pounds 

When  a  uniformly  distributed  load  occurs,  a  more  or  less  approxi- 
mate method  must  be  employed.     The  uniform  load  is  divided  into 


ENGINEERING  FOR  ARCHITECTS 


a  number  of  small  concentrated  loads.  Assuming  that  there  is  a 
uniformly  distributed  load  of  100  pounds  per  foot  over  a  beam  16 
feet  in  length,  it  is  necessary  to  divide  this  load  into  smaller  units  — 
100  pounds  —  and  treat  the  problem  as  though  there  were  sixteen 
concentrated  loads,  as  shown  in  Fig.  64.  The  diagrams  are  drawn 


FIGURE  64 

as  shown,  the  pole  distance  being  taken  as  2,000  pounds,  gf  in 
this  case  measures  1.6  feet  and  the  maximum  bending  moment  will 
be  1.6x2,000=  3,200  foot-pounds.  This  can  be  checked  by  the 
formula  M  =  \Wl.  |  x  1,600  X  16  =  3,200  foot-pounds. 

It  must  be  remembered  that  not  only  is  it  possible  to  discover 
the  maximum  bending  moment,  but  any  moment  can  be  found 
at  any  point  in  the  beam  by  this  method.  At  a  distance  4-feet 
6  inches  from  R\  the  distance  (bj)  scaled  on  the  diagram  is  1.29  feet> 


86        ENGINEERING  FOR  ARCHITECTS 

so   the   bending   moment   at   this   point  is   2,580  foot-pounds.     To 

check  this: 

+  800  pounds  X  4.5    feet  =  3,600  foot-pounds 

—  450  pounds  x  2.25  feet  =  1,012  foot-pounds 

M  =  2,588  foot-pounds 

These  problems  have  been  given  in  order  to  show  how  the  action 
of  forces  can  be  determined,  and  to  familiarize  the  architect  with 
graphical  methods.  These  cases  are  seldom  used  in  practice,  but  in 
some  instances  are  used  as  checks  to  locate  a  possible  error  in  calcula- 
tion. For  this  purpose  the  graphical  determination  of  moments  is 
very  valuable. 

The  principal  use  of  such  methods  is,  however,  in  determining 
the  stresses  in  roof  trusses.  Trusses,  having  loads  due  to  the  roof 
construction,  have  to  be  designed  to  withstand  the  vertical  pressure 
of  such  loads.  There  are  also  wind  loads  that  are  considered  as 
coming  upon  the  roof  in  a  direction  perpendicular  to  the  upper 
chord  of  the  truss.  This  makes  the  use  of  such  methods  as  have 
been  outlined  in  this  chapter  of  great  value,  as  they  can  be  used  to 
determine  the  action  of  forces  in  any  direction  or  of  any  magnitude. 

In  the  next  chapter  the  design  of  roof  trusses  will  be  taken  up, 
but  to  understand  the  considerations  involved  it  will  be  necessary 
to  understand  at  least  the  most  important  of  the  methods  given 
above. 


CHAPTER   XI 

Simple  Truss  Problems.     Fan  truss.     Fink  truss.     French  truss.     Wind  load 
diagram,  both  ends  of  truss  fixed. 

IN   the  last  chapter  the  methods  of  determining   the   action   of 
forces  by  means  of  graphical  methods  was  considered.     The  most 
practical  use  to  which  these  methods  are  put  is  that  of  designing 
roof  trusses.     The  average  architect    regards  the  design  of  trusses 
as  a  difficult  matter,  but  there  is  no  need  of  this. 

A  truss  may  be  represented  by  a  simple  triangle  as  shown  in 
Fig.  65   by  the  lines  XT,   TZ  and  ZX.     Suppose  a  force  of  100 


Ibs 


FIGURE  65 

pounds  is  applied  at  T.  This  force  is  carried  to  the  supports  RI 
and  R»  through  TX  and  TZ,  and  these  two  members  are  in  com- 
pression. If  the  member  XZ  were  not  there,  there  would  be  an 
outward  thrust  at  each  reaction.  XZ  acts  as  a  tie,  holding  the  ends 
of  the  truss  in  position.  This  member  is  in  tension. 

The  problem  that  a  designer  has  to  decide  is  the  exact  amount 
of  compression  or  tension  that  occurs  in  any  particular  member. 


ENGINEERING  FOR  ARCHITECTS 

In  order  to  do  this  a  system  of  lettering  is  employed  known   as 
"Bow's  Notation." 

Note  the  positions  of  A,  B,  C  and  D  in  Fig.  650.  The  line  XT 
now  is  between  A  and  D.  The  line  TZ  is  between  #  and  D.  The 
loo-pound  load  falls  between  A  and  B.  This  means  that  instead 
of  using  the  letters  at  the  ends  of  a  line  to  designate  the  line,  the 
letters  on  either  side  of  it  are  used.  In  other  words  XT  becomes 
DA,  the  force  of  100  pounds  becomes  AB,  TZ  becomes  BD,  and  XZ 
is  known  as  CD.  In  Fig.  65^  it  will  be  seen  that  X,  T  and  Z  are 
done  away  with  altogether.  The  reactions  also  are  lettered  in  this 
manner.  RI  is  CA  and  R2  is  BC.  The  diagram  in  which  these  let- 
ters are  shown  is  known  as  the  truss  diagram.  To  the  right  of  the 
truss  diagram,  Fig.  65,  is  another  known  as  the  stress  diagram. 

When  the  stress  diagram  is  drawn  the  use  of  Bow's  Notation 
becomes  apparent.  The  three  known  forces  are  the  two  reactions 
and  the  downward  load.  On  the  stress  diagram  lay  off  ab  parallel 
to  AB  and  equal  to  100  pounds.  In  order  to  find  the  stresses  in 
DA  and  BD  draw  a  line  through  a  parallel  to  AD,  and  one  through 
b  parallel  to  BD,  and  the  point  of  intersection  of  these  lines  must  be 
d.  By  measuring  da  or  bd  the  magnitude  of  the  stresses  in  the 
compression  members  can  be  determined.  The  amount  of  ten- 
sion in  the  lower  member  can  be  found  by  drawing  a  line  through 
d  parallel  to  DC  and  a  line  through  a  which  is  parallel  to  CA.  This 
last  line  will  coincide  with  the  line  ab.  The  point  of  intersection  of 
the  vertical  with  the  horizontal  line  must  be  c.  By  measuring  cd 
the  stress  in  CD  is  determined. 

Once  c  is  established  the  amount  of  weight  coming  upon  the 
supports  is  known.  It  has  already  been  pointed  out  that  CA  and 
^i  are  the  same,  and  this  is  true  of  BC  and  R%.  ca  and  be  are  both 
given  in  the  stress  diagram.  It  will  be  noted  that  be  equals  ca  and 
that  each  equals  one-half  of  ab.  As  the  load  AB  is  placed  directly 
in  the  center  of  the  span,  it  is  plain  that  each  reaction  must  equal 
one-half  the  load. 

In  all  the  work  in  which  graphical  methods  are  employed  the 
lines  that  give  the  magnitude  of  stresses,  in  the  stress  diagram,  must 
be  parallel  to  the  members  in  which  the  stresses  exist  in  the  truss 
diagram.  In  other  words,  ab  is  parallel  to  AB,  be  to  BC,  and  cd  to  CD. 

All  trusses  are  not  as  easily  developed  as  the  one  given  above. 
The  principles  of  determining  the  stresses  in  the  members  are,  how- 
ever, exactly  the  same  in  all.  The  truss  shown  in  Fig.  66  is  known 


ENGINEERING  FOR  ARCHITECTS 


89 


as  a  "Fan  Truss"  and  can  be  used  to  span  over  openings  of  from  20 
to  35  feet.  The  points  numbered  i,  2,  3,  4  and  5  are  known  as  panel 
points.,  and  the  load  upon  the  truss  is  generally  considered  as  acting 
as  concentrated  loads  at  these  points.  For  purposes  of  demonstra- 
tion let  it  be  assumed  that  a  force  of  100  pounds  acts  at  i,  a  force  of 
200  pounds  at  2,  another  at  3,  and  so  on  as  shown  in  the  figure.  The 
truss  diagram  is  lettered  as  shown  according  to  Bow's  Notation. 

The  next  step  is  to  lay  off  the  stress  diagram,  starting  with  the 
forces  already  known.  These  forces  are  BC,  CD,  DDf,  etc.  The 
reactions  are  also  known  and  so  the  points  a  can  be  established,  ab 
being  one-half  the  length  of  W . 

The  unknown  forces  are  found  by  considering  each  joint  sepa- 
rately. Start  at  the  first  panel  point  and  read  the  forces  in  a  clock- 


FIGURE  66 

wise  direction.  This  means  that  the  order  in  which  the  forces  are 
read  is  that  indicated  by  the  arrow  (Fig.  66)  and  corresponds  to  the 
direction  taken  by  the  hands  of  a  clock.  In  other  words  the  forces 
acting  at  the  first  panel  point  are  read  as  follows:  AB,  BC,  CE,  EA. 
Looking  at  the  stress  diagram,  ab  and  be  are  known  ce  is  not  known 
but  its  direction  is  parallel  to  the  upper  chord  of  the  truss  and  the 
point  c  is  known.  Draw  a  line  through  c  parallel  to  CE  and  continue 
it  indefinitely.  Neither  is  ea  known,  but  a  is  established,  and  it  is 
obvious  that  if  ea  is  parallel  to  EA  it  is  a  horizontal  line,  so  by  produc- 
ing a  horizontal  line  through  a  until  it  intersects  the  line  through  c,  the 
point  e  is  found,  e  is  common  to  both  ce  and  ea  and  therefore,  the 
intersection  of  the  two  lines  gives  the  point  that  was  to  be  found. 
By  scaling  the  length  of  ce  and  ea  the  magnitude  of  the  stresses  in 
the  corresponding  members  of  the  truss  are  found. 

In  order  to  tell  whether  these  stresses  are  compressive  or  tensile 
forces  a  very  simple  process  is  employed.  As  the  forces  are  laid 
off  on  the  stress  diagram,  ab  is  read  up.  This  stands  to  reason  as 


ENGINEERING  FOR  ARCHITECTS 


AE  is  the  left  reaction  —  R\  —  and  therefore  acts  up.  be  is  down, 
ce  acts  down  and  to  the  left.  If  an  arrow  is  placed  on  CE  indicating 
the  direction  in  which  ce  is  read,  this  arrow  must  point  toward  the 
panel  point  around  which  the  forces  are  taken.  This  means 
that  CE  is  in  compression.  When  the  direction  is  toward  the  joint 
the  stress  is  compressive. 

ea  is   read  from  left  to  right.     An   arrow  head   indicating  this 
direction  on  EA  is  away  from  the  panel  point.     This  means  that 


FIGURE  67 

EA  is  in  tension.  When  the  direction  is  away  from  the  joint  the  stress 
is  a  tensile  stress.  The  fact  that  CE  is  in  compression  and  EA  is  in 
tension  is  apparent  even  without  this  demonstration. 

Next  consider  the  stresses  and  the  force  acting  at  the  second 
panel  point,  ec  is  known,  cd  is  also  laid  off  on  the  stress  diagram. 
The  only  things  known  about  df  are  the  point  d  and  the  direction. 
Draw  through  d  a  line  parallel  to  DF.  The  next  member  to  be  found 
is  fe.  The  point  e  is  established  and  through  it  draw  a  line  parallel 
to  FE.  The  intersection  of  the  two  last  lines  gives  the  point  /. 

The  next  two  stresses  to  be  found  are  those  in  FG  and  GA.  In 
order  to  do  this  consider  the  stresses  around  the  joint  in  the  lower 


ENGINEERING  FOR  ARCHITECTS  91 

member,  ae  is  known,  and  so  is  ef,  but  only  the  directions  of  fg 
and  ga  are  given.  Through/  and  a  draw  lines  parallel  to  FG  and  GA, 
respectively,  and  the  intersection  establishes  the  point  g. 

By  the  methods  already  outlined  the  stresses  in  all  the  other 
members  of  the  truss  can  be  determined.  The  complete  diagram  is 
shown  in  Fig.  66. 

To  find  whether  EF  and  FG  are  in  tension  or  compression  it  will 
be  necessary  to  follow  around  the  stress  diagram  in  the  same  manner 
as  before.  Reading  around  the  joint  in  the  lower  chord,  ae  is  read 
away  from  the  joint  and  is  a  tensile  force;  ef  is  toward  the  joint  and 
is,  therefore,  a  compressive  force;  fg  and  ga  are  both  read  away  from 
the  joint  and  so  both  FG  and  GA  are  subject  to  pulls  of  the  magnitudes 
given  in  the  stress  diagram. 

A  diagrammatic  representation  of  a  Fink  truss  is  shown  in  Fig.  67 
and  as  this  type  of  truss  is  somewhat  different  from  many  others 
it  will  be  considered  next.  All  the  loads  are  considered  as  acting  at 
the  panel  points  as  in  the  truss  already  considered.  The  downward 
loads  are  BC,  CD,  DE,  EF,  FF',  etc.  The  upper  reactions  are  BA 
and  AB.  As  the  truss  is  symmetrically  loaded  the  reactions  will 
be  equal  and  each  will  equal  one-half  the  total  load.  The  point  a 
on  the  stress  diagram  falls  half  way  between  b  and  bf,  and  all  the 
other  loads  are  laid  off  in  exactly  the  same  manner  as  in  the  other 
examples. 

The  next  step  is  to  determine  the  stresses  in  the  members  at  the 
first  panel  point,  ab  is  known  and  so  is  be.  eg  and  ga  can  be  found 
in  the  same  manner  as  ce  and  ea  were  found  in  Fig.  66.  gc,  cd,  db 
and  bg  are  also  easily  found.  The  next  joint  is  in  the  lower  chord 
and  ag,  gh,  hi  and  ia  offer  no  particular  difficulties.  At  panel  point 
No.  3,  however,  difficulties  arise.  There  are  three  unknowns,  ek, 
kj,  and  ji,  and  without  special  information  it  will  be  impossible  to 
find  the  points  k  and  j  on  the  stress  diagram.  If  it  were  possible  to 
find  either  one  of  these  two  points,  the  other  can  be  found.  Now 
it  is  a  characteristic  of  the  Fink  truss  that  the  points  corresponding 
to  g,  b,  k  and  /  are  all  in  line  with  each  other  as  shown  in  the  stress 
diagram,  Fig.  67,  and  so,  by  continuing  gh  until  it  intersects  the  line 
through  e,  the  point  k  is  found.  This  method  can  be  proved  to  be 
correct,  but  the  proof  would  involve  the  architect  in  calculations 
more  or  less  complex,  and  which  for  all  practical  purposes  are  useless. 

Once  k  is  established,  the  point  j  is  found  by  drawing  a  line 
through  i  parallel  to  ij,  and  another  through  k  parallel  to  kj.  The 


ENGINEERING  FOR  ARCHITECTS 


intersection  of  these  two  lines  gives  the  desired  point.  It  is  then 
possible  to  read  the  stresses  in  ih,  hd,  de,  ek,  kj  and  ji.  The  whole 
process  depends  upon  rinding  the  point  k. 

All  other  stresses  can  be  found  without  any  particular  difficulty. 
This  is  not  only  true  of  this  particular  truss  but  of  all  trusses.  The 
methods  given  above  are  those  used  in  designing  the  most  compli- 
cated arched  trusses  or  those  having  unsymmetrical  loads.  A  con- 
dition in  which  there  are  loads  on  the  lower  member  is  shown  in  Fig. 


FIGURE  68 

68,  and  the  only  new  problem  is  the  method  of  laying  off  these  loads. 
It  will  also  be  noted  that  the  members  AG  and  AI  are  not  horizontal. 
This  form  of  Fink  truss  is  sometimes  given  the  name  of  French  truss, 
and  is  useful  when  headroom  is  required. 

The  loads  are  laid  off,  to  begin  with,  in  much  the  same  manner 
as  before,  until  the  point  b'  is  reached.  The  next  force,  b'af,  is  the 
reaction  R%.  In  all  previous  examples  this  has  been  laid  off  on  the 
stress  diagram  equal  to  one-half  the  distance  lib' .  In  this  case  b'a' 
is  equal  to  more  than  one-half  bbf.  As  the  loads  are  symmetrical, 


ENGINEERING  FOR   ARCHITECTS  93 

R2  is  equal  to  one-half  the  total  load  or  43  tons.  Note  that  the  3 
tons  on  the  lower  chord  is  figured  in  this  reaction.  The  point  a' 
falls  just  under/,  a'x  is  the  next  force,  and  xa  and  ab  bring  one  back 
to  the  starting  point  b  again. 

Once  these  forces  are  laid  off  in  order  and  according  to  a  very 
logical  arrangement,  the  work  of  finding  the  stresses  in  the  members 
of  the  truss  is  undertaken,  ab,  be  and  cd  are  laid  off  as  usual  but  it 
will  be  noted  that  ga  is  not  horizontal.  The  only  horizontal  line  in 
the  stress  diagram  is  yx  and  TX  is  the  only  horizontal  line  in  the 
truss  diagram.  In  the  diagrams  for  both  Fink  trusses  it  is  well  to 
note  that  /  and/  and  y  are  all  in  line.  This  relation  serves  as  a  check. 

The  stress  diagrams  that  have  been  given  so  far  have  been  de- 
veloped to  give  the  stresses  in  all  the  members  of  the  truss.  There 
is  no  need  of  this  in  actual  practice,  and  usually  only  one-half  the 
diagram  is  shown.  It  will  be  noted  that  a'g'  is  the  same  length  as 
ag,  and  the  same  relation  holds  good  throughout  the  entire 
diagram. 

All  the  forces  that  have  been  mentioned  so  far  have  been  down- 
ward loads.  In  case  wind  loads  are  to  be  considered  a  new  element 
enters  into  the  design.  The  usual  method  is  to  determine  the  stresses 
due  to  the  vertical  forces  and  then  draw  a  second  stress  diagram  in 
which  the  wind  loads  only  are  taken  into  account. 

These  wind  loads  are  considered  as  acting  at  the  panel  points  in 
exactly  the  same  manner  as  the  dead  loads.  The  method  of  deter- 
mining them  will  be  considered  later.  The  direction  of  these  loads 
is  always  taken  as  perpendicular  to  the  upper  chord  of  the  truss. 
This  condition  is  shown  in  Fig.  69.  The  determination  of  the  stresses 
in  the  members  depends  upon  the  manner  in  which  the  ends  of  the 
truss  are  held.  In  mill  construction  both  ends  of  a  truss  are  fixed 
and  this  is  true  of  many  trusses  that  are  used  in  building  construc- 
tion. When  one  end  of  the  truss  is  anchored  to  a  masonry  wall  the 
other  end  may  rest  upon  an  iron  plate  to  allow  for  expansion  and 
contraction.  When  a  truss  is  over  70  feet  in  length  the  free  end  is 
often  placed  upon  rollers.  The  need  of  this  is  apparent  if  one  con- 
siders the  difference  in  temperature  between  summer  and  winter 
and  the  expansion  that  may  take  place  in  70  feet  of  steel. 

When  both  ends  are  fixed  the  reactions  act  in  a  direction  opposite 
to  the  wind  loads,  that  is,  perpendicular  to  the  upper  chord  of  the 
truss.  The  stress  diagram  is  laid  off  in  much  the  same  manner  as 
those  abeady  considered  except  the  loads  are  not  vertical.  There 


94 


ENGINEERING  FOR   ARCHITECTS 


is  only  need  for  one  diagram,  for  all  forces  and  stresses  are  the  same 
whether  the  wind  blows  from  the  right  or  left. 

The  important  difference  between  this  condition  and  those  given 
above  is  that  it  is  necesary  to  determine  the  reactions.  In  the  case 
of  a  simple  beam  the  method  of  finding  R2  was  to  determine  the  total 
downward  moment  around  RI  caused  by  the  loads,  and  to  divide  this 
by  the  span.  Exactly  the  same  process  is  employed  in  reference  to 
the  truss.  The  total  wind  pressure  is  4,000  pounds  which  may  be 


considered  as  acting  at  panel  point  No.  3.  Its  lever  arm  is  11.5  feet. 
The  moment  caused  by  it  is  46,000  foot-pounds.  The  lever  arm  of 
RZ  around  RI  is  34.8  feet.  It  must  be  remembered  that  a  moment 
is  the  product  of  a  force  multiplied  by  the  perpendicular  distance 
from  the  center  of  moments.  RI  is  the  center  of  moments  in  this 
case.  jR2  multiplied  by  34.8  must  equal  46,000  foot-pounds.  So 
46,000  -T-  34.8  =  1,322  =  R2.  4,000  -  1,322  =  2,678  pounds  =  RI. 

Once  the  reactions  are  determined,  they  can  be  laid  off  on  the 
stress  diagram.  ^2  in  this  case  is  F'A  and  fa  laid  off  on  the  line  bf 
places  a  just  below  e.  ab  gives  the  left  reaction,  which  equals  2,678 


ENGINEERING  FOR  ARCHITECTS  95 

pounds.  We  now  have  be,  cd,  de,  ef,  ff',f'a  and  db.  The  remainder 
of  the  stress  diagram  is  laid  off  in  the  same  manner  as  those  given 
before.  As  the  points  g  and  b  coincide  with  each  other,  the  member 
GH  in  the  truss  diagram  has  no  stress  in  it  and  might  as  well  have 
been  omitted,  except  that  it  acts  as  a  support  to  keep  the  lower 
members  of  the  truss  from  sagging.  It  will  also  be  found  that  there 
will  be  no  stress  in  m'l',  I'k',  and  like  members  on  the  right-hand  side 
of  the  truss  when  the  wind  is  from  the  left.  This  is  shown  by  the 
fact  that  m'  and  I'  coincide  with  each  other. 

In  order  to  make  use  of  the  diagrams  already  drawn  it  is  neces- 
sary to  make  a  table.  In  this  table  there  must  be  spaces  for  the  mem- 
bers of  the  truss  and  opposite  each  member  is  recorded  the  stress  in 
it,  first,  when  the  load  is  vertical,  second,  when  the  wind  blows  from 
the  right,  third,  when  the  wind  blows  from  the  left,  and  the  sum 
of  those  stresses  that  give  the  greatest  total  stress.  Each  stress  must 
be  marked  plus  or  minus  to  indicate  if  it  is  a  tensile  or  compressive 
stress.  The  actual  design  of  the  members  is  then  undertaken. 

So  far  no  attention  has  been  paid  to  the  condition  in  which  one 
end  of  the  truss  is  free.  This  problem  will  be  dealt  with  in  the  next 
chapter  as  well  as  a  few  considerations  dealing  with  the  design  of 
the  truss. 


CHAPTER  XII 

Wind  Load  Diagram — One  end  free.     Tabulating  of  stresses.     Cantilever  Trusses. 

IN  order  to  establish  the  reactions  in  case  one  end  of  a  truss  is  fixed 
and  the  other  end  free,  the  same  method  is  employed  as  in  the 
case  of  a  simple  beam,  but,  because  the  wind  load  is  considered  as 
acting  in  a  direction  perpendicular  to  the  upper  chord  of  the  truss, 
and  the  right  reaction,  —  R2  —  which  is  supporting  the  free  end, 
acts  upward  in  a  vertical  direction,  there  is  a  tendency  to  consider 
that  this  process  is  very  difficult. 

Fig.  70  is  the  same  as  Fig.  69  except  the  right  end  of  the  truss  is 
considered  as  resting  on  rollers  or  on  an  iron  plate  over  which  it  can 


FIGURE  70 

slide.  The  advantage  of  this  is  that  allowance  is  made  for  expansion 
or  contraction  due  to  changes  of  temperature.  If  there  is  no  restric- 
tion of  the  action  of  the  stress  in  a  horizontal  direction,  the  only 
action  that  can  be  developed  at  R2  is  an  upward  one.  It  is  therefore 
necessary  to  determine  what  this  upward  action  is. 

The  moment  around  RI  caused  by  the  wind  is  exactly  the  same  as 
in  the  case  where  both  ends  of  the  truss  are  fixed.     The  load  is  still 


ENGINEERING  FOR  ARCHITECTS        97 

4,000  pounds  and  is  considered  as  acting  at  panel  point  No.  3.  The 
lever  arm  is  11.5  feet.  The  moment  is  therefore  the  same  as  before 
—  46,000  foot-pounds.  To  withstand  this  moment  an  opposite 
moment  must  be  caused  by  ^2- 

As  $2  in  this  case  acts  upwards,  the  perpendicular  distance  from 
^i  to  R2  is  40  feet.  The  equation  which  gives  the  magnitude  of  R2 
is  46,000  foot-pounds  =  R2  X  40  feet,  or,  R2  =  46,000  -f-  40  =  1,150 
pounds. 

Because  the  condition  is  applied  to  a  truss,  and,  because  the 
forces  are  spoken  of  as  wind  loads  and  reactions,  there  is  a  tendency 
to  make  a  symmetry  of  this  example.  If  the  same  conditions  were 
applied  to  a  compound  lever,  such  as  found  in  automobiles,  the 


/,/so 


FIGURE  71 

mystery  may  be  explained  in  a  more  simple  manner.  In  Fig.  71  such 
a  lever  is  shown  which  is  supposed  to  be  pivoted  at  RI,  around  which 
point  it  can  swing  freely  and  a  load  of  4,000  pounds  is  supposed  to  be 
applied  at  X  in  a  direction  perpendicular  to  R\X.  This  force  would 
tend  to  swing  the  lever  around  jRi,  a  moment  of  46,000  foot-pounds 
having  been  created.  But  a  force  is  exerted  upward  at  R2  which 
tends  to  hold  the  lever  in  its  position.  This  force  has  a  lever  arm  of 
40  feet.  How  much  force  must  be  exerted  at  RZ?  This  is  exactly 
the  same  example  as  given  in  the  preceding  case. 

Once  RZ  is  determined  the  stress  diagram  is  drawn  be,  cd,  de,  ef, 
and  f  are  laid  off  as  in  Fig.  69,  but  fa  offers  a  new  problem.  As 
has  been  said  F'A,  the  right  reaction,  is  a  vertical  force  and  so  fa 
must  be  drawn  upward.  The  length  of  the  line  is  determined  by 
the  magnitude  of  F'A  or  1,150  pounds.  The  architect  may  have 
noticed  that  RI  has  not  yet  been  determined,  but  all  that  it  is  neces- 
sary to  do  in  order  to  establish  both  the  direction  and  magnitude  of 
the  left  reaction  is  to  connect  a  and  b.  The  line  ab  gives  the  neces- 


ENGINEERING  FOR  ARCHITECTS 


sary  information  about  AB,  or  jRi.  Once  all  the  external  forces 
are  plotted,  the  next  step  is  the  determination  of  the  stresses  in  the 
members.  This  is  done  in  exactly  the  same  manner  as  in  Fig.  69. 
It  will  be  noticed  that  eg,  di,  ck,  fm,  etc.,  are  of  the  same  magnitude 
as  in  the  case  where  both  ends  of  the  truss  are  fixed.  The  only 
members  in  which  the  stresses  are  different  are  those  in  the  lower 
chord  GA,  HA,  JA,  and  LA. 

When  both  ends  of  the  truss  were  fixed  only  one  stress  diagram 
was  necessary.     When  the  wind  blows  from  the  right  the  stresses 


FIGURE  72 

in  the  members  on  the  right  side  of  the  truss  would  equal  those  in 
the  left  side  when  the  conditions  were  those  shown  in  Fig.  69.  With 
one  end  fixed  and  the  other  free  two  wind  stress  diagrams  are  neces- 
sary. This  is  due  to  the  fact  that  RI  and  Rz  are  different  when  the 
wind  blows  from  opposite  sides  of  the  roof. 

The  method  of  finding  RZ  in  the  case  where  the  wind  blows  from 
the  right  needs  but  little  explanation.  Fig.  72  is  like  that  of  the  case 
where  the  wind  is  from  the  left,  the  only  difference  being  that  the 
lever  arm  of  the  wind  load  is  longer.  This  follows  from  the  fact 
that  the  direction  of  the  wind  load  is  different,  and  the  perpendicular 


ENGINEERING  FOR  ARCHITECTS 


99 


distance  from  RI  to  the  4,ooo-pound  load  is  23.25  feet.  The  moment 
is  4,000  pounds  x  23.25  feet  =  93,000  foot-pounds. 

The  lever  arm  of  R%  is  the  same  as  before,  40  feet,  so  R%  must  equal 
93,000  -=-  40  =  2,325  pounds. 

If  it  is  desired  to  show  how  this  condition  is  represented  by  a 
compound  lever,  note  the  diagram  shown  in  Fig.  73.  The. lever 
arms  and  loads  are  represented  in  the  most  simple  manner. 

Once  R%  is  determined  the  stress  diagram  is  plotted  —  Fig.  72. 
It  will  be  noticed  that  the  loads  start  with  FF'  and  read  in  the  stress 
diagram  in  the  following  order:  ff,  f'c',  c'd',  d'c',  c'b',  b'a,  and  of. 


Z.3Z5 


The  left  reaction  —  RI  —  is  given  by  the  last  named  line  —  of — and 
its  direction  is  shown  on  the  truss  diagram. 

In  the  two  wind-load  diagrams  and  the  dead-load  diagram  shown 
in  Fig.  74  all  the  stresses  that  may  occur  in  the  truss  are  shown.  It 
now  remains  for  the  architect  to  scale  the  lines  which  show  the 
stresses  and  to  tabulate  them  as  shown  in  Fig.  75.  This  table  is  self 
explanatory.  The  stresses  are  recorded  in  the  total  column  and  the 
signs  denoting  them  represent  the  kind  of  stress  in  each  member. 
These  stresses  are  the  largest  that  will  occur.  Each  total  is  not  the 
sum  of  all  the  stresses,  but  represents  the  sum  of  the  dead-load  stress 
plus  the  greatest  wind-load  stress. 

To  design  the  members  of  the  truss  it  is  necessary  to  know  the 
kind  of  stress  that  occurs  in  each.  If  the  member  is  in  tension  all 
that  is  necessary  to  do  is  to  divide  the  stress  by  16,000  pounds  and 
the  number  of  square  inches  necessary  in  each  particular  member  is 
given. 

On  the  other  hand  the  compressive  stress  tend  to  buckle  mem- 
bers in  which  they  occur  and  these  members  must  be  designed  as 
small  columns.  The  formula  S  =  15,200  —  $8l/r  is  used  in  the  same 


100 


ENGINEERING  FOR  ARCHITECTS 


manner  as  given  in  a  previous  chapter.  For  the  purpose  of  sim- 
plifying the  design  of  the  truss  the  upper  chord  is  made  of  angles  that 
will  withstand  the  stress  occurring  in  CG  although  the  stresses  in 
Z)7,  EK  and  FM  are  smaller  than  this.  The  same  is  true  of  the  angles 
in  the  lower  chord,  the  stress  that  governs  the  design  is  found  in  GA. 
Although  smaller  stresses  may  occur  in  members  on  the  left  side  or 


FIGURE  74 

right  side  of  the  truss,  as  the  case  may  be,  both  sides  are  made 
alike. 

It  is  not  necessary  to  design  all  the  details  of  the  truss  as  the 
steel  contractor  will  furnish  shop  drawings  of  these.  It  is  neces- 
sary, however,  to  give  the  stresses  that  occur  and  to  know  the  num- 
ber of  rivets  and  size  of  members  required. 

It  will  be  found  in  some  tension  members  that  the  angles  which 
will  be  strong  enough  to  take  up  the  stress  will  be  too  small  to  be 
riveted.  In  this  case  angles  having  legs  of  least  2j  inches  must  be 
considered  the  smallest  that  are  practical. 

Because  of  the  need  of  only  small  members  where  tension  exists, 
trusses  are  designed  with  tie  rods  to  withstand  the  tensile  stresses. 


ENGINEERING  FOR  '4%CStritCr$  101 


Such  trusses  are  called  "pin  trusses."  The  compression  members 
are  made  of  deck  beams  or  angles  or  of  some  other  cast  or  rolled 
shapes. 

Often  in  galleries  in  theaters  there  is  need  of  cantilever  trusses. 
These  offer  two  new  considerations  for  the  architect  to  deal  with. 
First  there  is  the  proposition  of  unsymmetrical  loads  and,  second, 
the  fact  that  either  RI  or  R%  may  not  act  at  the  end,  but  somewhere 
near  the  center  of  the  truss. 

This  last  fact  gives  rise  to  the  necessity  of  determining  the  reac- 
tions and  the  method  used  is  that  of  finding  the  reactions  of  a  simple 
cantilever  beam.  For  the  purpose  of  this  chapter  the  truss  shown  in 
Fig.  76  will  explain  the  method  of  solving  such  a  problem.  Loads 
of  1,000  pounds  each  are  assumed  as  acting  at  the  panel  points  1-6, 


MtMTiE.*L 

£>EAE>  LOAD 

WlNt>l-OAT>U 

WiNj?  LOAP  15. 

TbTAt-S 

0  G* 

+  21,000 

-t-  3,  6  0  0 

+  2.300 

+24  6OO 

X)   1 

•f  ig.OOO 

t    3,  £  00 

+    2,300 

+21  ,200 

E  K. 

•*•!  5.000 

t   2.600 

,_t  2/300 

+  (7,feOo 

r  M 

+  IZ,000 

•f   2.    1  O  0 

t  Z.300 

-t)4,  /oo 

G   A 

—  16,500 

-5,000 

-    /,  000 

-13,500 

H  /^ 

-  18.500 

-  5>ooo 

-  /,  ooo 

-  |3,5oc 

J   A 

-  15,  750 

-4,000 

-   I  ,  ooo 

-  1  9,  750 

L.  A> 

-  13,  000 

-  3,000 

—  1,000 

-  1  6,  ooo 

H  J 

+  9,  ooo 

•+•   1  ,  I  75 

O 

•+    I.I75" 

I   J 

—  )>5oo 

600 

0 

-  Z.i  ob 

J    K 

-t-^,  ooo 

+  1,500 

0 

+  5:500 

K  L. 

-  3,000 

-    1,200 

0 

-4,200 

I-  M 

4  5  000 

-»-  £,  ooo 

0 

t  7,ooa 

M  M' 

—  9,  000 

-  ^  ,«oo 

-  1,  f°° 

-10,  e>oo 

FIGURE  75 

and  loads  of  500  pounds  each  acting  at  panel  points  o  and  7.  R%  acts 
under  panel  point  4.  Taking  moments  around  panel  point  o  the 
total  downward  moment  is  given  as: 

500  pounds  X  o  feet  =  ooo  foot-pounds. 
1,000  "  x  10  "  =  10,000 
i, ooo  "  x  20  "  =  20,000 
i, ooo  "  x  30  "  =  30,000 
i, ooo  "  x  40  "  =  40,000 
1,000  "  x  50  "  =  50,000 

1,000          "          X  60     "      =  60,000 

500       "       x  70    "    =  35>°QQ 

245,000  foot-pounds. 


7,000  pounds 
245,000  foot-pounds  4-  40  feet  =  6,125  pounds. 
7,000-6,125=      875 


102 


MGINE'ERTNG  FOR  ARCHITECTS 


It  will  be  noticed  that  almost  all  the  entire  load  comes  upon  R2, 
and  that  had  there  been  a  greater  load  or  a  greater  projection  beyond 
R2  there  might  have  been  a  negative  or  downward  reaction  at  RI. 
The  force  ab,  be,  cd,  de,  <?/,  fg,  gh,  and  ho  are  laid  off.  The  upward 
reaction  —  RZ  —  o'o  causes  o  to  fall  just  below  a  and  the  left  reaction 


o 
-40-0" 


o 

-30-0' 


H 


FIGURE  76 

—  RI  —  closes  the  force  diagram.     The  stresses  are  laid  off  in  the 
same  manner  as  in  the  preceding  diagrams. 

To  find  the  forces  one  may  start  at  either  end  of  the  truss.  Start- 
ing at  the  right  end  and  reading  this  first  the  forces  and  stresses  can 
be  laid  off  as  follows:  hof,  o'y  and  yh.  There  will  be  no  stress  in  XT 
and  the  stresses  around  panel  point  6  can  next  be  found.  No  diffi- 
culty will  be  experienced  until  R%  is  reached  and  there  need  be  no 
trouble  here  if  it  is  remembered  that  the  force  of  6,125  pounds  — 
o'o  —  is  read  upward. 

If  there  should  be  a  case  in  which  there  is  a  negative  moment 
at  £1  as  is  shown  in  Fig.  77  the  reactions  are  determined  as  before. 
The  three  i,ooo-pound  loads  and  5oo-pound  load  will  tend  to  bend 
the  truss  around  R%  and  to  find  the  magnitude  of  this  reaction  take 
moments  around  RI. 


ENGINEERING  FOR  ARCHITECTS 


103 


i,ooo  pounds  X    8  feet  =    8,000  foot-pounds. 
1,000       "       X  14    "    =  14,000 

1,000          "          X  20     "      =  20,000 

700       "       X  26    "    =  I3>°o° 
3,500  pounds  55>ooo  foot-pounds. 

55,000-^  8  =  6,875  pounds. 
3,500-  6,875  =   -  3.375  =  ^i- 

The  fact  that  RI  is  a  minus  quantity  shows  that  RI  acts  in  an 
opposite  direction  to  R%.     In  laying  off  the  forces  it  will  be  noticed 


FIGURE  77 

that  there  is  only  one  upward   force  —  R2.     Starting  at   the  first 
joint  ab  is  plotted,     bg  is  a  horizontal  line  and  ga  closes  the  diagram. 

It  will  be  found  that  there  is  no  stress  in  GH  as  this  is  simply  a 
bracing  member.  At  2  gh  is  known  and  be.  ci  can  be  drawn  through 
c  and  ih  can  be  laid  off. 

Laying  off  the  forces  and  stresses  around  the  lower  joint,  hi  is 
known,  ij  is  drawn  indefinitely  through  i  in  a  direction  parallel 
to  IJ.  The  joint  /  is  known  on  the  stress  diagram  and  so  jf  can  be 
plotted.  The  intersection  of  these  two  lines  gives  the  point  /.  fa 
is  the  upward  reaction  —  R^  —  and  ah  has  already  been  drawn. 
Around  panel  point  3  ji,  ic  and  cd  are  already  established,  dk 


io4  ENGINEERING  FOR  ARCHITECTS 

and  kj  are  easily  found.  The  stresses  in  other  members  can  be  found 
without  any  particular  difficulty. 

In  this  book  the  considerations  taken  up  have  been  just  those 
with  which  the  architect  comes  in  contact.  There  are  cases  in  which 
the  engineering  requirements  of  an  undertaking  are  such  that  special 
knowledge  is  necessary  to  solve  the  problems  that  come  up.  In  this 
case  it  is  always  necessary  for  the  architect  to  call  upon  an  engineer 
to  design  the  steel. 

If,  however,  the  architect  has  found  in  this  book  such  informa- 
tion as  will  enable  him,  with  the  use  of  a  steel  handbook,  to  design 
the  frames  for  simple  architectural  structures,  the  object  for  which 
the  book  was  written  has  been  accomplished. 


UNIVERSITY  OF  CALIFORNIA  LIBRA! 
This  book  is  DUE  on  the  last  date  stamped  below. 

^illli&se^ 


m 


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MAY  19  1948 


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